2. Composite Systems and States: Tensor Products, Density Matrices, Partial Traces
This document discusses how to describe multiple quantum systems as a single composite system, and how to mathematically express quantum states that contain incomplete information. Tensor product is the language used to combine the spaces of each system, and density matrix is a universal state representation that encompasses both pure and mixed states. Partial trace is a key tool used when examining only a part of the entire system.
1. Fundamental Concepts
Tensor Product: A mathematical construction used to treat two or more quantum systems as a single composite system. When the Hilbert spaces of each system are \(H_A\) and \(H_B\), the state space of the entire system becomes the tensor product space \(H_A \otimes H_B\). The inner product in this space is naturally defined as the product of the inner products in each subspace: \(\langle x_1\otimes y_1, x_2\otimes y_2\rangle = \langle x_1,x_2\rangle_A \langle y_1,y_2\rangle_B\).
Detailed Explanation: Why is the Tensor Product Needed?
The tensor product provides a way to express the combination of all possible states each system can have. If we have coin A (heads/tails) and coin B (heads/tails), the state of the entire system becomes four cases: (A heads, B heads), (A heads, B tails), (A tails, B heads), (A tails, B tails). Similarly, the dimension of the Hilbert space is also multiplied (\(\dim(H_A \otimes H_B) = \dim(H_A) \times \dim(H_B)\)).
In particular, quantum mechanics goes beyond simple combinations to include entanglement states, which are non-classical phenomena that can only be defined on the tensor product structure.
Density Operator (Density Matrix): The most general mathematical object used to describe quantum states. It is defined as an operator \(\rho\) satisfying the following two conditions:
- Positive semidefinite: \(\rho \ge 0\), i.e., for all \(|\psi\rangle\), \(\langle\psi|\rho|\psi\rangle \ge 0\).
- Unit trace: \(\mathrm{Tr}(\rho)=1\). Detailed Explanation: Why is the Density Matrix Needed? (Distinction Between Pure and Mixed States)
The state vector \(|\psi\rangle\) describes only the pure state where the system’s information is perfectly known. This is fundamentally different from a state that is ‘50% probability of \(|0\rangle\), 50% probability of \(|1\rangle\)’. These two represent the difference between ‘quantum superposition’ and ‘classical ignorance’.
Pure State (Coherent Superposition): This is a state like \(|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\). This is a single, perfect state where the two states \(|0\rangle\) and \(|1\rangle\) have a clear phase relationship (coherence) and are ‘simultaneously’ superimposed. Although the measurement results split into 50% probabilities, the state before measurement is singular.
Mixed State (Incoherent Mixture): This is a state that is ‘50% probability of \(|0\rangle\), 50% probability of \(|1\rangle\)’. This represents classical probability (ignorance) where the system is actually either \(|0\rangle\) or \(|1\rangle\), but we don’t know which. There is absolutely no coherence (phase relationship) between the two states.
Mathematical Example: The density matrix clearly distinguishes these two.
Pure State \(\rho_{\text{pure}} = |\psi\rangle\langle\psi| = \left( \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \right) \left( \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \end{pmatrix} \right) = \frac{1}{2} \begin{pmatrix} 1 & \mathbf{1} \\ \mathbf{1} & 1 \end{pmatrix}\)
Mixed State \(\rho_{\text{mix}} = 0.5 |0\rangle\langle0| + 0.5 |1\rangle\langle1| = \frac{1}{2} \begin{pmatrix} 1 & \mathbf{0} \\ \mathbf{0} & 1 \end{pmatrix}\)
The pure state has off-diagonal terms (non-zero) indicating coherence (except for basis states like \(|0\rangle\)), whereas the mixed state has these off-diagonal terms as zero. A state like \(\rho_{\text{mix}}\) cannot be expressed with a single state vector \(|\psi\rangle\). The density matrix is a generalized tool that consistently describes both pure and mixed states. 👨🔬
Partial Trace: This is an operation that discards information about one subsystem (e.g., B) from the combined system \(\rho_{AB}\) and retrieves only the state \(\rho_A\) of the remaining subsystem (A). It is the quantum mechanical counterpart of obtaining the marginal probability distribution \(P(X) = \sum_Y P(X,Y)\) from the joint probability distribution \(P(X,Y)\) in probability theory. > Detailed Explanation: Partial Trace and Entanglement - “Perfect Whole” and “Imperfect Parts” > > Partial trace reveals a key feature of entanglement. At this point, the expression “information is perfect/imperfect” is based on purity (\(\mathrm{Tr}(\rho^2)\)) rather than the total sum of probabilities (\(\mathrm{Tr}(\rho)=1\), which always holds). > > * Perfect information (pure state): \(\mathrm{Tr}(\rho^2)=1\). > Why “perfect”?: Because we are 100% certain that the system’s state is a single state vector \(|\psi\rangle\). (e.g., we exactly know that the system is in the state \(\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\)). There is no classical ignorance here. > > * Imperfect information (mixed state): \(\mathrm{Tr}(\rho^2)<1\). > Why “imperfect”?: Because we cannot specify the system’s state with a single vector. Our knowledge remains in a statistical ensemble such as “50% probability for \(|0\rangle\) and 50% probability for \(|1\rangle\)”. This means there is classical ignorance about whether the system is actually in one of the two states. > > An entangled state (e.g., Bell state \(\rho_{AB}=|\Phi^+\rangle\langle\Phi^+|\)) is a pure state for the total system (\(\rho_{AB}\)) (\(\mathrm{Tr}(\rho_{AB}^2)=1\)). That is, “information about the whole is perfect”. (We are 100% certain that the system is in the single state \(|\Phi^+\rangle\).) > > However, if we take a partial trace of one side (B) and obtain the state of A, \(\rho_A = \mathrm{Tr}_B(\rho_{AB})\), it becomes \(\rho_A = \frac{1}{2}I\) (a completely mixed state). The purity of this state is \(\mathrm{Tr}(\rho_A^2) = 1/2 < 1\). > > In other words, “information about the part becomes imperfect”. This means that someone who only observes A can only describe the state of A as a statistical ensemble \(\frac{1}{2}I\) (i.e., they know nothing about A). Information about A is not possessed solely by A but exists only within the “relationship between A and B”, and partial trace makes information incomplete by discarding that relationship and focusing only on A.
Positive Operator: An operator \(A\) is called a positive operator if it satisfies \(\langle\psi|A|\psi\rangle \ge 0\) for all vectors \(|\psi\rangle\). This is equivalent to stating that all eigenvalues of the operator are non-negative. The density operator must be a positive operator because it is physically connected to the concept of “probability”.
2. Symbols and Key Relations
- Tensor Product and Inner Product:
- When \(H_A, H_B\) have orthonormal bases \(\{|i\rangle_A\}, \{|j\rangle_B\}\) respectively, the orthonormal basis of \(H_A \otimes H_B\) is given by \(\{|i\rangle_A \otimes |j\rangle_B\}\) or simply \(\{|ij\rangle\}\).
- Tensor product of operators: \((A \otimes B)(|a\rangle \otimes |b\rangle) = (A|a\rangle) \otimes (B|b\rangle)\).
- Density operator \(\rho\):
- Purity: The degree of mixing of the state is measured by the value of \(\mathrm{Tr}(\rho^2)\).
- Pure state: \(\rho=|\psi\rangle\langle\psi| \implies \mathrm{Tr}(\rho^2)=1\).
- Mixed state: \(\rho=\sum_i p_i |\psi_i\rangle\langle\psi_i| \implies \mathrm{Tr}(\rho^2) < 1\).
- Expectation value: The expectation value of an observable \(A\) is calculated as \(\langle A\rangle = \mathrm{Tr}(\rho A)\).
💡 Why is it called a “Density” Matrix? (Relation to Purity)
The name “Density Matrix” originates from the similarity to the probability density function in classical statistical mechanics. A probability density function indicates how concentrated the probability is that the system exists in a particular state.
- Pure state (Pure State) is a state where the system’s information is concentrated 100% in a single quantum state. It corresponds to the state with the highest ‘concentration’ of information.
- Mixed state (Mixed State) is a state where the information is distributed across multiple quantum states, i.e., a diluted state.
Therefore, the term “density” is an intuitive expression indicating how concentrated (high purity) or spread out (low purity) the quantum information is in the state space. Purity (
Tr(ρ²)) serves as an indicator measuring this concentration of information. - Purity: The degree of mixing of the state is measured by the value of \(\mathrm{Tr}(\rho^2)\).
- Partial trace \(\mathrm{Tr}_B\):
- Core definition: For a simple tensor product operator, it is defined as \(\mathrm{Tr}_B(A \otimes B) = A \cdot \mathrm{Tr}(B)\), and this is linearly extended to all operators.
- Matrix representation: The matrix elements of \(\rho_A\) are calculated as \((\rho_A)_{ij} = \sum_k (\rho_{AB})_{ik,jk}\). Here, \(ik, jk\) denote the indices of the combined system.
- Quantum channel (CPTP map):
- Physically possible changes in quantum states (quantum operations) are described by a Completely Positive and Trace-Preserving (CPTP) linear map \(\Phi\).
- Kraus representation (Kraus Representation): Every CPTP map can be expressed in the form \(\Phi(\rho) = \sum_k K_k \rho K_k^\dagger\), where the Kraus operators \(K_k\) satisfy \(\sum_k K_k^\dagger K_k = \mathbf{1}\) (trace-preserving condition).
3. Easy Examples (Examples with Deeper Insight)
- Tensor Product Example 1: Basis States of Two Qubits
- \(|0\rangle_A = \begin{pmatrix}1\\0\end{pmatrix}, |1\rangle_B = \begin{pmatrix}0\\1\end{pmatrix}\) when,
- \(|01\rangle = |0\rangle_A \otimes |1\rangle_B = \begin{pmatrix}1\\0\end{pmatrix} \otimes \begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}1\cdot\begin{pmatrix}0\\1\end{pmatrix} \\ 0\cdot\begin{pmatrix}0\\1\end{pmatrix}\end{pmatrix} = \begin{pmatrix}0\\1\\0\\0\end{pmatrix}\)
- Similarly, \(|00\rangle \to (1,0,0,0)^{\mathsf T}\), \(|10\rangle \to (0,0,1,0)^{\mathsf T}\), \(|11\rangle \to (0,0,0,1)^{\mathsf T}\) form the standard basis of the 4-dimensional space. Orthogonality can be easily verified as \(\langle 01|10\rangle = (0,1,0,0) \cdot (0,0,1,0)^{\mathsf T} = 0\).
- Density Matrix Example 1: Pure State vs. Mixed State
- Pure State: \(|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\). \(\rho=|\psi\rangle\langle\psi| = \frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix}\). \(\rho^2 = \frac{1}{4}\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}1&1\\1&1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&2\\2&2\end{pmatrix} = \rho\). \(\mathrm{Tr}(\rho^2) = \mathrm{Tr}(\rho)=1\).
- Mixed State: 50% probability of \(|0\rangle\), 50% probability of \(|1\rangle\). \(\rho=\frac{1}{2}|0\rangle\langle 0|+\frac{1}{2}|1\rangle\langle 1| = \frac{1}{2}\begin{pmatrix}1&0\\0&1\end{pmatrix} = \frac{\mathbf{1}}{2}\). \(\rho^2 = \frac{1}{4}\mathbf{1}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{4}\mathrm{Tr}(\mathbf{1})=\frac{2}{4}=\frac{1}{2} < 1\).
💡 Purity of \(\mathrm{Tr}(\rho^2)\)
The eigenvalues of the density matrix \(\rho\) can be interpreted as the probabilities \(\{p_i\}\) of being in each eigenstate. \(\mathrm{Tr}(\rho^2) = \sum_i p_i^2\).
- Pure State has a 100% probability of being in a specific state, so the eigenvalues are \(\{1, 0, \dots, 0\}\). Therefore, \(\sum p_i^2 = 1^2+0^2+\dots=1\).
- Mixed State has probabilities distributed across multiple states (\(\sum p_i = 1, p_i<1\)), so \(\sum p_i^2 < 1\) (e.g., \(0.5^2+0.5^2=0.5\)). The purity is minimized in the most mixed state (maximum entropy).
- Partial Trace Example 1: Calculation of Separable State (Product State)
- Consider a pure state without entanglement \(|\psi\rangle_{AB} = |0\rangle_A \otimes |+\rangle_B = |0\rangle_A \otimes \frac{1}{\sqrt{2}}(|0\rangle_B + |1\rangle_B) = \frac{1}{\sqrt{2}}(|00\rangle + |01\rangle)\).
- The total density matrix is \(\rho_{AB} = |\psi\rangle\langle\psi| = \frac{1}{2}(|00\rangle + |01\rangle)(\langle 00| + \langle 01|)\).
(Basis order: \(|00\rangle, |01\rangle, |10\rangle, |11\rangle\))
\(\rho_{AB} = \frac{1}{2}\begin{pmatrix}1&1&0&0\\1&1&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}\).
- Let’s compute \(\rho_A\) by taking the partial trace over the B system.
\(\rho_A = \mathrm{Tr}_B(\rho_{AB}) = \sum_{k=0,1} \langle k|_B \rho_{AB} |k\rangle_B = \langle 0|_B \rho_{AB} |0\rangle_B + \langle 1|_B \rho_{AB} |1\rangle_B\)
- \(\langle 0|_B \rho_{AB} |0\rangle_B\) extracts the submatrix corresponding to B being 0 (1st and 3rd rows/columns).
\(\frac{1}{2}\begin{pmatrix}1&0\\0&0\end{pmatrix}\)
- \(\langle 1|_B \rho_{AB} |1\rangle_B\) extracts the submatrix corresponding to B being 1 (2nd and 4th rows/columns).
\(\frac{1}{2}\begin{pmatrix}1&0\\0&0\end{pmatrix}\)
- Adding these two gives \(\rho_A\).
\(\rho_A = \frac{1}{2}\begin{pmatrix}1&0\\0&0\end{pmatrix} + \frac{1}{2}\begin{pmatrix}1&0\\0&0\end{pmatrix} = \begin{pmatrix}1&0\\0&0\end{pmatrix} = |0\rangle\langle 0|_A\).
💡 Interpretation of Example 1
The state \(\rho_A\) of A is a pure state with purity \(\mathrm{Tr}(\rho_A^2)=1\). This shows that in the case of a separable state without entanglement, the partial trace simply ignores the other system (\(|+\rangle_B\)) and returns the original state of A (\(|0\rangle_A\)). This intuitive result sharply contrasts with the next example. - Partial Trace Example 2: Reduction of Entangled State (Bell State) - Let’s examine the density matrix of the entangled state \(|\Phi^+\rangle=\tfrac{1}{\sqrt{2}}(|00\rangle+|11\rangle)\). \(\rho_{AB}=|\Phi^+\rangle\langle\Phi^+| = \frac{1}{2}(|00\rangle+|11\rangle)(\langle 00|+\langle 11|) = \frac{1}{2}\begin{pmatrix}1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1\end{pmatrix}\). - Let’s compute \(\rho_A\) by taking the partial trace over B. \(\rho_A = \mathrm{Tr}_B(\rho_{AB}) = \langle 0|_B \rho_{AB} |0\rangle_B + \langle 1|_B \rho_{AB} |1\rangle_B\) - Extracting the part where B=0 (1st and 3rd rows/columns): \(\frac{1}{2}\begin{pmatrix}1&0\\0&0\end{pmatrix}\) - Extracting the part where B=1 (2nd and 4th rows/columns): \(\frac{1}{2}\begin{pmatrix}0&0\\0&1\end{pmatrix}\) - Adding these two together. \(\rho_A = \frac{1}{2}\begin{pmatrix}1&0\\0&0\end{pmatrix} + \frac{1}{2}\begin{pmatrix}0&0\\0&1\end{pmatrix} = \frac{1}{2}\begin{pmatrix}1&0\\0&1\end{pmatrix} = \frac{\mathbf{1}}{2}\).
> **💡 Remarkable Implication of This Result**
>
> The entire AB system is in a completely known **pure state** ($\mathrm{Tr}(\rho_{AB}^2)=1$). However, if we consider only system A, its state $\rho_A$ becomes the most disordered **maximally mixed state** ($\mathrm{Tr}(\rho_A^2)=1/2$). The measurement results on A are perfectly random. But the moment we know the measurement result on B, the state of A is immediately determined. This is the non-classical characteristic of entanglement.- Positive Operator Example 2: Square Form Expression
- A positive operator \(A\) is equivalent to being expressible in the form \(A=B^\dagger B\).
- This is obvious because \(\langle\psi|A|\psi\rangle = \langle\psi|B^\dagger B|\psi\rangle = \langle B\psi|B\psi\rangle = \|B\psi\|^2 \ge 0\), analogous to how for a complex number \(z\), \(z^*z = |z|^2 \ge 0\). This shows that a positive operator has properties similar to a “square”.
4. Practice Problems
Basic Concept Check
- Tensor Product Calculation: Explicitly calculate the tensor product \(A \otimes B\) of the two operators \(A=\sigma_x, B=\sigma_z\) as a 4x4 matrix.
- Tensor Product Basis: Verify that the norm of the two-qubit state \(|\psi\rangle=\tfrac{1}{2}(|00\rangle+i|01\rangle-|10\rangle-i|11\rangle)\) is 1, and determine whether this state is a separable state (product state) of the form \(|\psi\rangle = |\alpha\rangle_A \otimes |\beta\rangle_B\).
- Density Matrix Construction: When a qubit is in the state \(|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\) with probability \(\frac{3}{4}\) and in the state \(|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)\) with probability \(\frac{1}{4}\), find the density matrix \(\rho\) of this system.
- Pure/Mixed State Discrimination: Calculate the purity \(\mathrm{Tr}(\rho^2)\) of the density matrix \(\rho\) obtained in problem 3 and determine whether it is a pure or mixed state.
- Expectation Value Calculation: Calculate the expectation value \(\langle \sigma_z\rangle\) of the observable \(\sigma_z\) for the state \(\rho\) in problem 3.
Partial Trace and Entanglement
- Verification of Partial Trace Definition: Prove that \(\mathrm{Tr}_A(\rho_{AB}) = \rho_B\) for a separable state where \(\rho_{AB} = \rho_A \otimes \rho_B\).
- Bell State Reduction: Write the density matrix \(\rho_{AB}\) for the entangled state \(|\Psi^-\rangle=\tfrac{1}{\sqrt{2}}(|01\rangle-|10\rangle)\) and calculate the reduced density matrix \(\rho_A=\mathrm{Tr}_B(\rho_{AB})\) for A.
- Entanglement Determination: For the density matrix \(\rho_{AB}\) of the following state, calculate the partial trace \(\rho_A\) and determine whether the original state is entangled by its purity: \(|\psi\rangle = \cos\theta|00\rangle + \sin\theta|11\rangle\).
- Partial Trace and Expectation Value: For a general composite state \(\rho_{AB}\) and an observable \(A_0\) acting only on system A, prove that calculating the expectation value of \(A_0 \otimes \mathbf{1}_B\) over the entire system is equivalent to calculating the expectation value of \(A_0\) over the reduced state \(\rho_A\). That is, \(\mathrm{Tr}_{AB}(\rho_{AB}(A_0 \otimes \mathbf{1}_B)) = \mathrm{Tr}_A(\rho_A A_0)\).
Positive Operators and Quantum Channels
- Determine if a Positive Operator: Determine whether the matrices \(A=\begin{pmatrix}2&-i\\i&1\end{pmatrix}\), \(B=\begin{pmatrix}1&2\\2&1\end{pmatrix}\) are positive operators. (Hint: Check the eigenvalues)
- Kraus Operator: The bit-flip channel is described by Kraus operators \(K_0=\sqrt{1-p}\,\mathbf{1}, K_1=\sqrt{p}\,\sigma_x\). Verify the condition \(\sum_k K_k^\dagger K_k=\mathbf{1}\), and calculate the state \(\Phi(\rho)\) after the initial state \(\rho=|0\rangle\langle 0|\) passes through this channel.
- Bloch Sphere Representation: Any single-qubit density matrix can be written as \(\rho=\tfrac{1}{2}(\mathbf{1}+\vec{r}\cdot\vec{\sigma})\). Show that \(\mathrm{Tr}(\rho^2) = \frac{1}{2}(1+\|\vec{r}\|^2)\), and explain how pure and mixed states are represented on the Bloch sphere using this.
5. Explanation
\(A \otimes B = \begin{pmatrix}0&1\\1&0\end{pmatrix} \otimes \begin{pmatrix}1&0\\0&-1\end{pmatrix} = \begin{pmatrix}0\cdot B & 1\cdot B \\ 1\cdot B & 0\cdot B \end{pmatrix} = \begin{pmatrix}0&0&1&0\\0&0&0&-1\\1&0&0&0\\0&-1&0&0\end{pmatrix}\).
\(\|\psi\|^2 = (\frac{1}{2})^2 (1^2+|i|^2+(-1)^2+|-i|^2) = \frac{1}{4}(1+1+1+1)=1\). Rearranging the state gives \(|\psi\rangle = \frac{1}{2}(|0\rangle(1|0\rangle+i|1\rangle) -|1\rangle(1|0\rangle+i|1\rangle)) = \frac{1}{2}(|0\rangle-|1\rangle) \otimes (|0\rangle+i|1\rangle)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
$^2 =
\[\begin{pmatrix}2&1\\1&2\end{pmatrix}\]\begin{pmatrix}2&1\1&2
\(A \otimes B = \begin{pmatrix}0&1\\1&0\end{pmatrix} \otimes \begin{pmatrix}1&0\\0&-1\end{pmatrix} = \begin{pmatrix}0\cdot B & 1\cdot B \\ 1\cdot B & 0\cdot B \end{pmatrix} = \begin{pmatrix}0&0&1&0\\0&0&0&-1\\1&0&0&0\\0&-1&0&0\end{pmatrix}\).
\(\|\psi\|^2 = (\frac{1}{2})^2 (1^2+|i|^2+(-1)^2+|-i|^2) = \frac{1}{4}(1+1+1+1)=1\). Rearranging the state gives \(|\psi\rangle = \frac{1}{2}(|0\rangle(1|0\rangle+i|1\rangle) -|1\rangle(1|0\rangle+i|1\rangle)) = \frac{1}{2}(|0\rangle-|1\rangle) \otimes (|0\rangle+i|1\rangle)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
\(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01| - |01\rangle\langle 10| - |10\rangle\langle 01| + |10\rangle\langle 10|)\). After normalization, \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\), so it is a separable state.
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
Okay, so I need to figure out what’s going on here. The user has provided a long list of equations, all seemingly related to quantum mechanics, specifically dealing with density matrices and expectation values. Let me try to parse through this step by step.
First, looking at the structure of the equations. Each entry from 1 to 100 seems to be a separate calculation. However, most of them are repetitive. For example, equations 1 to 100 are mostly the same, except for the numbering. The first equation is:
- \(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-|\)
Then, equation 2 is similar but with a different notation, and so on. Wait, no, actually looking at the equations, they all seem to be the same. Let me check again.
Wait, the user has written equations 1 through 100, but each equation is actually the same. For instance, equation 1 is:
\(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-|\)
Then equation 2 is:
\(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\)
Wait, but then equation 3 is:
\(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state.
Then equation 4 is:
\(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\).
Then equation 5 is:
\(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of density matrices, \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\).
Wait, but equations 1 to 100 are not all the same. Let me check again. The user has written equations from 1 to 100, but each equation seems to be a repetition of the same steps. However, looking at the numbering, equation 1 is the definition of the density matrix, equation 2 is the calculation of \(\rho^2\), equation 3 is the trace of \(\rho^2\), equation 4 is the expectation value of \(\sigma_z\), equation 5 is about partial trace, and then equations 6 to 100 repeat equations 1 to 5 again but with different numbers. Wait, that seems to be the case. For example, equation 6 is the same as equation 1 but numbered 6, equation 7 is the same as equation 2 but numbered 7, etc. So the user has written the same 5 equations 20 times, resulting in 100 equations. That’s a bit strange, but perhaps they are testing something or just made a mistake in formatting.
But regardless, the main content here is the calculation of the density matrix, its square, the trace of the square, the expectation value of \(\sigma_z\), and the partial trace. Let me focus on that.
First, the density matrix is given as \(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-|\). Here, \(|+\rangle\) and \(|-\rangle\) are the eigenstates of \(\sigma_x\), right? Because \(\sigma_x |+\rangle = |+\rangle\) and \(\sigma_x |-\rangle = -|-\rangle\). However, the user then calculates \(\rho^2\), which might be part of determining whether the state is mixed or pure. The trace of \(\rho^2\) for a pure state is 1, and for a mixed state, it’s less than 1. Here, they compute \(\mathrm{Tr}(\rho^2) = \frac{5}{8}\), which is less than 1, confirming it’s a mixed state.
Then, they calculate the expectation value of \(\sigma_z\), which comes out to zero. That makes sense because the density matrix is a statistical mixture of \(|+\rangle\) and \(|-\rangle\), which are eigenstates of \(\sigma_x\), not \(\sigma_z\). Therefore, the expectation value of \(\sigma_z\) should be zero, as there’s no preference for up or down in the z-direction.
The partial trace part (equation 5) seems to be about taking the partial trace over one subsystem, but the notation is a bit unclear. The equation is \(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). However, this seems incorrect because the partial trace over A of \(\rho_A \otimes \rho_B\) should be \(\rho_B\), since \(\rho_A\) is a density matrix on system A and \(\rho_B\) on system B. But the equation says that the trace of \(\rho_A\) times \(\rho_B\), but actually, the partial trace would just be \(\rho_B\). The user might have made a mistake here, or perhaps they’re using a different notation where \(\mathrm{Tr}_A\) is being applied to \(\rho_A \otimes \rho_B\), which is redundant because \(\rho_A \otimes \rho_B\) is already a composite state. Taking the partial trace over A would leave \(\rho_B\), which is correct. However, the equation written by the user is \(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). But \(\mathrm{Tr}(\rho_A)\) is 1 because it’s a density matrix, so this simplifies to \(\rho_B\). Therefore, that equation is correct in that sense.
But why is this part included? Maybe the user is trying to show that the partial trace of a product state is the state of the other subsystem, which is a basic property in quantum mechanics.
Now, putting this all together, the main points here are:
- The density matrix is a statistical mixture of \(|+\rangle\) and \(|-\rangle\) with probabilities 3/4 and 1/4.
- Calculating \(\rho^2\) and finding that \(\mathrm{Tr}(\rho^2) = 5/8 < 1\), confirming the state is mixed.
- The expectation value of \(\sigma_z\) is zero, indicating no bias in the z-direction.
- The partial trace of a product state gives the state of the other subsystem.
However, the repetition of the same equations 20 times seems unnecessary and might be a formatting error or a test. The user might have intended to present this once but mistakenly repeated it. Alternatively, they might be showing a process where each step is repeated multiple times, but that doesn’t make much sense here.
Another thing to note is the calculation of \(\rho^2\). Let’s verify that. Given \(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-|\), squaring this would be:
\[ \rho^2 = \left(\frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-|\right)^2 = \left(\frac{3}{4}\right)^2 |+\rangle\langle+| + \left(\frac{1}{4}\right)^2 |-\rangle\langle-| + \frac{3}{4} \cdot \frac{1}{4} (|+\rangle\langle+| |-\rangle\langle-| + |-\rangle\langle-| |+\rangle\langle+|)) \]
But since \(|+\rangle\langle+| |-\rangle\langle-| = |+\rangle\langle+|-\rangle\langle-| = 0\) because \(|+\rangle\) and \(|-\rangle\) are orthogonal. Therefore, the cross terms vanish, and we get:
\[ \rho^2 = \left(\frac{9}{16}\right)|+\rangle\langle+| + \left(\frac{1}{16}\right)|-\rangle\langle-| \]
So, the density matrix squared is just the same form but with coefficients squared. Now, the user then writes:
\[ \rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix} \]
Wait, this seems to be a different approach. Let me check the matrix representation. Let’s assume that \(|+\rangle\) and \(|-\rangle\) are represented as \(\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1\end{pmatrix}\) and \(\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -1\end{pmatrix}\), respectively. Then, the density matrix \(\rho\) would be:
\[ \rho = \frac{3}{4} \cdot \frac{1}{2} \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix} + \frac{1}{4} \cdot \frac{1}{2} \begin{pmatrix}1 & -1 \\ -1 & 1\end{pmatrix} = \frac{3}{8} \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix} + \frac{1}{8} \begin{pmatrix}1 & -1 \\ -1 & 1\end{pmatrix} \]
Adding these together:
\[ \left(\frac{3}{8} + \frac{1}{8}\right)\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} + \frac{3}{8} - \frac{1}{8} \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix} = \frac{1}{2}\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} + \frac{1}{4} \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix} \]
Which simplifies to:
\[ \begin{pmatrix}\frac{1}{2} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{2}\end{pmatrix} \]
So, \(\rho\) is represented as \(\begin{pmatrix}\frac{1}{2} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{2}\end{pmatrix}\). Then, \(\rho^2\) would be:
\[ \begin{pmatrix}\frac{1}{2} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{2}\end{pmatrix} \cdot \begin{pmatrix}\frac{1}{2} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{2}\end{pmatrix} \]
Calculating the matrix multiplication:
Top-left element: \(\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{4} + \frac{1}{16} = \frac{5}{16}\)
Top-right element: \(\frac{1}{2} \cdot \frac{1}{4} + \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} + \frac{1}{8} = \frac{1}{4}\)
Similarly, the bottom elements are the same due to symmetry:
\[ \rho^2 = \begin{pmatrix}\frac{5}{16} & \frac{1}{4} \\ \frac{1}{4} & \frac{5}{16}\end{pmatrix} = \frac{1}{16} \begin{pmatrix}5 & 4 \\ 4 & 5\end{pmatrix} \]
Which matches the user’s calculation. Then, the trace of \(\rho^2\) is \(\frac{5}{16} + \frac{5}{16} = \frac{10}{16} = \frac{5}{8}\), confirming the mixed state.
For the expectation value of \(\sigma_z\), the matrix representation of \(\sigma_z\) is \(\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\). Then, \(\rho \sigma_z\) is:
\[ \begin{pmatrix}\frac{1}{2} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{2}\end{pmatrix} \cdot \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix} = \begin{pmatrix}\frac{1}{2} & -\frac{1}{4} \\ \frac{1}{4} & -\frac{1}{2}\end{pmatrix} \]
Then, the trace of this matrix is \(\frac{1}{2} - \frac{1}{2} = 0\), which matches the user’s result.
The partial trace part is a bit confusing. If \(\rho_A \otimes \rho_B\) is a product state, then the partial trace over A would give \(\rho_B\), assuming that \(\rho_A\) is a density matrix on system A and \(\rho_B\) on system B. However, the user’s equation is written as \(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}(\rho_A) \cdot \rho_B\). Since \(\mathrm{Tr}(\rho_A) = 1\), this simplifies to \(\rho_B\), which is correct. However, the notation here might be a bit off because the partial trace is an operation, not a multiplication. The user might have intended to write \(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \rho_B\), which is accurate.
In summary, the user has correctly calculated the density matrix, its square, the trace of the square, the expectation value of \(\sigma_z\), and the partial trace. However, the repetition of the same equations 20 times is likely an error or a formatting issue. The main points are all correct and demonstrate understanding of mixed states, expectation values, and partial traces in quantum mechanics.
The provided equations demonstrate key concepts in quantum mechanics, particularly focusing on density matrices, mixed states, and expectation values. Here’s a structured breakdown of the analysis:
1. Density Matrix Definition
The density matrix is given as: \[ \rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| \] - \(|+\rangle\) and \(|-\rangle\) are eigenstates of \(\sigma_x\), representing a superposition of spin states along the x-axis. - This is a mixed state since it is a statistical mixture of two orthogonal states with probabilities \(3/4\) and \(1/4\).
2. Square of the Density Matrix
\[ \rho^2 = \left(\frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-|\right)^2 = \frac{9}{16}|+\rangle\langle+| + \frac{1}{16}|-\rangle\langle-| \] - The cross terms vanish because \(|+\rangle\) and \(|-\rangle\) are orthogonal. - Matrix representation (using \(\sigma_x\) eigenstates): \[ \rho = \begin{pmatrix}\frac{1}{2} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{2}\end{pmatrix}, \quad \rho^2 = \frac{1}{16}\begin{pmatrix}5 & 4 \\ 4 & 5\end{pmatrix} \]
3. Trace of \(\rho^2\)
\[ \mathrm{Tr}(\rho^2) = \frac{5}{16} + \frac{5}{16} = \frac{5}{8} < 1 \] - Confirms the state is mixed (pure states have \(\mathrm{Tr}(\rho^2) = 1\)).
4. Expectation Value of \(\sigma_z\)
\[ \langle \sigma_z \rangle = \mathrm{Tr}(\rho \sigma_z) = 0 \] - \(\sigma_z = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\), and the trace of \(\rho \sigma_z\) is zero. - This indicates no net polarization along the z-axis, consistent with the state being a mixture of \(|+\rangle\) and \(|-\rangle\) (which are symmetric with respect to \(\sigma_z\)).
5. Partial Trace
\[ \mathrm{Tr}_A(\rho_A \otimes \rho_B) = \rho_B \] - For a product state \(\rho_A \otimes \rho_B\), the partial trace over system A yields \(\rho_B\). - The user’s equation \(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}(\rho_A) \cdot \rho_B\) is correct since \(\mathrm{Tr}(\rho_A) = 1\).
Summary
- The calculations correctly demonstrate the properties of a mixed state, the use of density matrices, and the computation of expectation values.
- The repeated equations (likely a formatting error) do not affect the validity of the results.
- The analysis confirms the state is mixed, has zero \(\sigma_z\) polarization, and aligns with the mathematical formalism of quantum mechanics.
- \(A \otimes B = \begin{pmatrix}0&1\\1&0\end{pmatrix} \otimes \begin{pmatrix}1&0\\0&-1\end{pmatrix} = \begin{pmatrix}0\cdot B & 1\cdot B \\ 1\cdot B & 0\cdot B \end{pmatrix} = \begin{pmatrix}0&0&1&0\\0&0&0&-1\\1&0&0&0\\0&-1&0&0\end{pmatrix}\). 2. \(\|\psi\|^2 = (\frac{1}{2})^2 (1^2+|i|^2+(-1)^2+|-i|^2) = \frac{1}{4}(1+1+1+1)=1\). When rearranging the state \(|\psi\rangle = \frac{1}{2}(|0\rangle(1|0\rangle+i|1\rangle) -|1\rangle(1|0\rangle+i|1\rangle)) = \frac{1}{2}(|0\rangle-|1\rangle) \otimes (|0\rangle+i|1\rangle)\). After normalization \(|\alpha\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle), |\beta\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\) so it is a separable state. 3. \(\rho = \frac{3}{4}|+\rangle\langle+| + \frac{1}{4}|-\rangle\langle-| = \frac{3}{4}\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} + \frac{1}{4}\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \frac{1}{8}\begin{pmatrix}3+1&3-1\\3-1&3+1\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\). 4. \(\rho^2 = \frac{1}{16}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix} = \frac{1}{16}\begin{pmatrix}5&4\\4&5\end{pmatrix}\). \(\mathrm{Tr}(\rho^2) = \frac{1}{16}(5+5) = \frac{10}{16} = \frac{5}{8} < 1\). Therefore, it is a mixed state. 5. \(\langle \sigma_z\rangle = \mathrm{Tr}(\rho\sigma_z) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) = \mathrm{Tr}\left(\frac{1}{4}\begin{pmatrix}2&-1\\1&-2\end{pmatrix}\right) = \frac{1}{4}(2-2)=0\). 6. \(\mathrm{Tr}_A(\rho_A \otimes \rho_B) = \mathrm{Tr}( \rho_A ) \cdot \rho_B\). By the definition of the density matrix \(\mathrm{Tr}(\rho_A)=1\), so the result is \(\rho_B\). 7. \(\rho_{AB} = \frac{1}{2}(|01\rangle-|10\rangle)(\langle 01|-\langle 10|) = \frac{1}{2}(|01\rangle\langle 01|-|01\rangle\langle 10|-|10\rangle\langle 01|+|10\rangle\langle 10|)\). \(\rho_A = \mathrm{Tr}_B(\rho_{AB}) = \frac{1}{2}(\langle 1|\dots|1\rangle_B + \langle 0|\dots|0\rangle_B) = \frac{1}{2}(|0\rangle\langle 0| + |1\rangle\langle 1|) = \frac{\mathbf{1}}{2}\). 8. \(\rho_{AB} = (\cos\theta|00\rangle + \sin\theta|11\rangle)(\cos\theta\langle 00| + \sin\theta\langle 11|)\). \(\rho_A = \mathrm{Tr}_B(\rho_{AB}) = \cos^2\theta|0\rangle\langle 0| + \sin^2\theta|1\rangle\langle 1| = \begin{pmatrix}\cos^2\theta&0\\0&\sin^2\theta\end{pmatrix}\). \(\mathrm{Tr}(\rho_A^2) = \cos^4\theta + \sin^4\theta\). If \(\theta \in (0, \pi/2)\), this value is less than 1, so \(\rho_A\) is a mixed state, and therefore the original state is an entangled state. Only when \(\theta = 0\) or \(\pi/2\) does it become a separable state. 9. In the basis \(\{|i\rangle_A |j\rangle_B\}\), \(\mathrm{Tr}_{AB}(\rho_{AB}(A_0 \otimes \mathbf{1}_B)) = \sum_{i,j} \langle i|_A\langle j|_B \rho_{AB}(A_0 \otimes \mathbf{1}_B) |i\rangle_A|j\rangle_B\) \(= \sum_{i,j} \langle i|_A A_0 (\sum_k \langle k|_B \rho_{AB} |k\rangle_B) |i\rangle_A\). (Here, \(A_0\) is independent of \(B\)) \(\sum_k \langle k|_B \rho_{AB} |k\rangle_B\) is exactly the definition of \(\rho_A\), \(= \sum_i \langle i|_A A_0 \rho_A |i\rangle_A = \mathrm{Tr}_A(A_0 \rho_A) = \mathrm{Tr}_A(\rho_A A_0)\). 10. The eigenvalues of \(A\) are \(\det(A-\lambda I) = (2-\lambda)(1-\lambda)-1=0 \implies \lambda^2-3\lambda+1=0 \implies \lambda = \frac{3\pm\sqrt{5}}{2} > 0\). Therefore, it is a positive operator. The eigenvalues of \(B\) are \(\det(B-\lambda I) = (1-\lambda)^2-4=0 \implies \lambda=3, -1\). Since there is a negative eigenvalue, it is not a positive operator. 11. \(\sum K_k^\dagger K_k = (1-p)\mathbf{1}^\dagger\mathbf{1} + p\,\sigma_x^\dagger\sigma_x = (1-p)\mathbf{1} + p\,\mathbf{1} = \mathbf{1}\). \(\rho = \begin{pmatrix}1&0\\0&0\end{pmatrix}\). \(\Phi(\rho) = K_0\rho K_0^\dagger + K_1\rho K_1^\dagger = (1-p)\rho + p\,\sigma_x\rho\sigma_x = (1-p)\begin{pmatrix}1&0\\0&0\end{pmatrix} + p\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} = (1-p)\begin{pmatrix}1&0\\0&0\end{pmatrix} + p\begin{pmatrix}0&0\\0&1\end{pmatrix} = \begin{pmatrix}1-p&0\\0&p\end{pmatrix}\). This means that the bit is flipped with probability \(p\). 12. \(\rho^2 = \frac{1}{4}(\mathbf{1}+\vec{r}\cdot\vec{\sigma})^2 = \frac{1}{4}(\mathbf{1} + 2\vec{r}\cdot\vec{\sigma} + (\vec{r}\cdot\vec{\sigma})^2)\). \((\vec{r}\cdot\vec{\sigma})^2 = \sum_{i,j}r_i r_j \sigma_i \sigma_j = \sum_i r_i^2 \sigma_i^2 + \sum_{i\neq j} r_i r_j \sigma_i \sigma_j = \|\vec{r}\|^2\mathbf{1}\). (cross terms cancel)
Therefore \(\rho^2 = \frac{1}{4}((1+\|\vec{r}\|^2)\mathbf{1} + 2\vec{r}\cdot\vec{\sigma})\). \(\mathrm{Tr}(\rho^2) = \frac{1}{4}((1+\|\vec{r}\|^2)\mathrm{Tr}(\mathbf{1}) + 2\sum_i r_i \mathrm{Tr}(\sigma_i)) = \frac{1}{4}((1+\|\vec{r}\|^2)\cdot 2 + 0) = \frac{1}{2}(1+\|\vec{r}\|^2)\).
- Pure state: \(\mathrm{Tr}(\rho^2)=1 \iff \|\vec{r}\|^2=1 \iff \|\vec{r}\|=1\). i.e., points on the surface of the Bloch sphere.
- Mixed state: \(\mathrm{Tr}(\rho^2)<1 \iff \|\vec{r}\|<1\). i.e., points in the interior of the Bloch sphere. The maximally mixed state (\(\mathbf{1}/2\)) corresponds to the origin \(\vec{r}=0\).