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1 Why Density Matrices Appear
1.1 Physical Motivation
In standard quantum mechanics, we describe systems using state vectors |ω→. This works perfectly for pure states—completely specified quantum states. However, in real phys- ical situations, we often encounter:
- Entangled composite systems: When system A is entangled with system B, we cannot fully describe A alone with a single state vector, even if the joint system AB is pure.
- Statistical mixtures: A system might be in state |ω 1 → with probability p 1 or in state |ω 2 → with probability p 2. This is fundamentally di!erent from a quantum superposition.
- Subsystems of entangled states: If we only have access to subsystem A (and cannot measure B), the state of A appears mixed from our perspective, even though AB is pure.
1.2 The Logic Flow
The fundamental physical picture is:
- Complete description (full system): Pure state |ω (^) AB → in H (^) A ↑ H (^) B
- Partial description (subsystem A only): Must use density operator εA to account for our ignorance of B
- Measurement predictions: For observables on A alone, we compute expectation values using Tr(εA O (^) A )
- Entropy: Quantifies how “mixed” a state is—how much information is hidden
The density operator is the natural mathematical tool for describing states when we have incomplete information about the quantum system.
2 Definition and Properties
2.1 Definition of Density Matrix
A density operator (or density matrix) ε is an operator on Hilbert space that de- scribes a quantum state. It can be written as:
ε =
k
p (^) k |ω (^) k →↓ω (^) k | (1)
where p (^) k ↔ 0 are probabilities and
k p^ k^ = 1. Physical interpretation: The system is in state |ω (^) k → with probability p (^) k.
2.2 Key Mathematical Properties
Every valid density operator must satisfy:
- Hermitian: ε†^ = ε
- All eigenvalues are real (measurable quantities)
- Trace equals one: Tr(ε) = 1
- Sum of all probabilities equals 1
- If ε =
k p^ k^ |ω^ k^ →↓ω^ k^ |, then Tr(ε) =^
k p^ k^ = 1
- Positive semi-definite: ↓ϑ| ε |ϑ→ ↔ 0 for all |ϑ→
- All eigenvalues ϖi ↔ 0
- This is equivalent to saying ε has a spectral decomposition
2.3 Spectral Decomposition
Any density matrix can be diagonalized:
ε =
k
ϖ (^) k |ϱk →↓ϱ (^) k | (2)
where:
- ϖk are eigenvalues (probabilities): ϖ (^) k ↔ 0,
k ϖ^ k^ = 1
- |ϱk → are orthonormal eigenvectors
- Physical meaning: We can think of the system as being in eigenstate |ϱk → with probability ϖ (^) k
4 The Partial Trace
4.1 Definition and Intuition
The partial trace is an operation that “traces out” (removes) information about one subsystem while keeping the description of the other subsystem. For a bipartite system with Hilbert space H (^) A ↑ H (^) B , the partial trace of operator O ↘ H (^) A ↑ H (^) B over subsystem B is:
Tr (^) B (O) =
∑^ d^ B
j=
(I (^) A ↑ ↓b (^) j |)O(I (^) A ↑ |b (^) j →) (8)
where {|b (^) j →} d j^ B=1 is any orthonormal basis of H (^) B. Key property: This definition is basis-independent—the result doesn’t depend on which basis we choose.
4.2 Why Partial Trace Works
Consider a product state |ϑ (^) A → ↑ |ω (^) B →:
Tr (^) B (|ϑ (^) A →↓ϑ (^) A | ↑ |ω (^) B →↓ω (^) B |) = |ϑ (^) A →↓ϑ (^) A | · Tr(|ω (^) B →↓ω (^) B |) = |ϑ (^) A →↓ϑ (^) A | (9) since Tr(|ω (^) B →↓ω (^) B |) = 1. For a general product operator O (^) A ↑ O (^) B :
Tr (^) B (O (^) A ↑ O (^) B ) = OA · Tr(OB ) (10) This shows that tracing out B e!ectively “collapses” the B-part while leaving A un- changed.
5 Reduced Density Matrices
5.1 Definition
For a bipartite system in state |ω (^) AB →, or more generally with density operator ε (^) AB , the reduced density matrix of subsystem A is:
εA = Tr (^) B (ε (^) AB ) (11) Similarly, for subsystem B:
εB = Tr (^) A (ε (^) AB ) (12)
5.2 Computing the Partial Trace: Explicit Formula
If ε (^) AB acts on H (^) A ↑ H (^) B and {|b (^) j →} is an orthonormal basis of H (^) B :
ε (^) A = Tr (^) B (ε (^) AB ) =
j
(I (^) A ↑ ↓b (^) j |)εAB (I (^) A ↑ |b (^) j →) (13)
Matrix form: If we expand εAB in the product basis {|a (^) i , b (^) j →} = {|a (^) i → ↑ |b (^) j →}:
ε (^) AB =
i,i →^ ,j,j →
c (^) ij,i →^ j →^ |ai , bj → ↓ai →^ , bj →^ | (14)
Then:
ε (^) A =
i,i →^ ,j
c (^) ij,i → (^) j |ai → ↓ai → | (15)
Notice: We only sum over states with matching B indices (j = j →^ ).
5.3 Example: Two Qubits
Consider a system of two qubits with state:
ε (^) AB =
(Here, |ij→ = |i→ (^) A ↑ |j→ (^) B ) To find ε (^) A , use the basis {| 0 → (^) B , | 1 → (^) B }: For j = 0: (I (^) A ↑ ↓ 0 | (^) B )ε (^) AB (I (^) A ↑ | 0 → (^) B ) =
| 0 → ↓ 0 | A +
| 1 → ↓ 1 | A (17)
For j = 1: (I (^) A ↑ ↓ 1 | (^) B )ε (^) AB (I (^) A ↑ | 1 → (^) B ) =
| 0 → ↓ 0 | A (18)
Therefore:
ε (^) A =
| 0 → ↓ 0 | A +
| 1 → ↓ 1 | A +
| 0 → ↓ 0 | A =
| 0 → ↓ 0 | A +
| 1 → ↓ 1 | A (19)
6 Purification and Schmidt Decomposition
6.1 The Purification Problem
Question: Given a density operator ε (^) A on H (^) A , can we always find a larger Hilbert space H (^) B and a pure state |ω (^) AB → ↘ HA ↑ H (^) B such that:
εA = Tr (^) B (|ω (^) AB →↓ω (^) AB |) (21) Answer: Yes! This process is called purification.
6.2 Canonical Purification
Given the spectral decomposition εA =
k p^ k^ |ϱk^ →↓ϱ^ k^ |^ with^ p^ k^ >^ 0, construct an ancilla (auxiliary) system B with orthonormal basis {|ςk →} and define:
|ω (^) AB → =
k
p (^) k |ϱk → (^) A ↑ |ς (^) k → (^) B (22)
Then:
Tr (^) B (|ω (^) AB →↓ω (^) AB |) =
k
p (^) k |ϱk →↓ϱ (^) k | (^) A = εA (23)
Key properties:
- The minimal dimension of the ancilla is rank(εA ) (the number of non-zero eigen- values)
- Purifications are unique up to unitary transformations on the ancilla
6.3 Schmidt Decomposition
For a pure state |ω (^) AB → ↘ H (^) A ↑ H (^) B , the Schmidt decomposition expresses it as:
|ω (^) AB → =
∑^ r
k=
s (^) k |ϱk → (^) A ↑ |ς (^) k → (^) B (24)
where:
- sk ↔ 0 are Schmidt coe”cients with
k s^ (^2) k = 1
- {|ϱk → (^) A } and {|ςk → (^) B } are orthonormal sets
- r is the Schmidt rank (minimum number of terms needed)
6.4 Reduced States and Schmidt Decomposition
From the Schmidt decomposition, the reduced density matrices are:
ε (^) A = Tr (^) B (|ω (^) AB →↓ω (^) AB |) =
k
s (^2) k |ϱk →↓ϱ (^) k | (^) A (25)
ε (^) B = Tr (^) A (|ω (^) AB →↓ω (^) AB |) =
k
s (^2) k |ς (^) k →↓ς (^) k | (^) B (26)
Important: Both reduced density matrices have the same spectrum {s^2 k }.
6.5 Example: Bell State
Consider the maximally entangled state:
∣∣ ” +^
This is already in Schmidt form with:
- Schmidt rank: r = 2
- Schmidt coe#cients: s 1 = s 2 = ↑^12
The reduced states are:
ε (^) A =
I
ε (^) B =
I
Both are maximally mixed states—complete ignorance about each qubit individu- ally.
8 Example Worked Problems
8.1 Problem 1: Pure Product State
Question: For a two-qubit system in state |ω→ = | 0 → (^) A | 1 → (^) B , find the reduced density matrices. Solution: The joint density operator is:
εAB = |ω→↓ω| = | 01 → ↓ 01 | (32)
To find ε (^) A , trace over B using basis {| 0 → (^) B , | 1 → (^) B }:
ε (^) A =
j
(I (^) A ↑ ↓j| (^) B ) | 01 → ↓ 01 | (I (^) A ↑ |j→ (^) B ) (33)
For j = 0: ↓ 0 | (^) B 01 → = 0 For j = 1: ↓ 1 | (^) B 01 → = ↓ 1 | 1 → = 1
ε (^) A = | 0 → (^) A ↓ 0 | (^) A = | 0 → ↓ 0 | (34) This is a pure state since (εA ) 2 = εA. Similarly, εB = | 1 → ↓ 1 | is also pure. Key insight: Product states have pure reduced states for both subsystems.
8.2 Problem 2: Entangled State
Question: For a two-qubit system in the Bell state |” +^ → = ↑^12 (| 00 → + | 11 →), compute:
- The joint density operator
- The reduced density matrix of qubit A
- The von Neumann entropy of ε (^) A
Solution: Part 1: Joint density operator
εAB =
” +^
” +^
Part 2: Reduced density matrix for A using basis {| 0 → (^) B , | 1 → (^) B }: For j = 0: (I (^) A ↑ ↓ 0 | (^) B )ε (^) AB (I (^) A ↑ | 0 → (^) B ) =
| 0 → ↓ 0 | A (36)
For j = 1: (I (^) A ↑ ↓ 1 | (^) B )ε (^) AB (I (^) A ↑ | 1 → (^) B ) =
| 1 → ↓ 1 | A (37)
Therefore: εA =
| 0 → ↓ 0 | A +
| 1 → ↓ 1 | A =
I
Part 3: Von Neumann entropy Eigenvalues of εA : { 1 / 2 , 1 / 2 }
S(ε (^) A ) = ⇐
ln(1/2) ⇐
ln(1/2) = ln 2 (39) Key insight: The joint state is pure (S(ε (^) AB ) = 0), but the reduced state is maximally mixed (S(ε (^) A ) = ln 2). This is the hallmark of entanglement.
9.2 Question 2: Pure vs Mixed States (10 marks)
Question: Consider two density operators for a qubit:
- ε 1 = |+→ ↓+| where |+→ = ↑^12 (| 0 → + | 1 →)
- ε 2 = 12 | 0 → ↓ 0 | + 12 | 1 → ↓ 1 | Part a) [5 marks] For each state, calculate Tr(ε^2 ) and determine whether it is pure or mixed. Part b) [5 marks] Write out ε 1 and ε 2 as 2 ⇑ 2 matrices in the computational basis {| 0 → , | 1 →}. Explain physically why ε 2 is mixed while ε 1 is pure, without referring to the trace.
Complete Solution
Part a) Answer: For ε 1 = |+→ ↓+|: First, calculate ε^21 :
ε 21 = (|+→ ↓+|)(|+→ ↓+|) = |+→ ↓+|+→ ↓+| = |+→ ↓+| = ε 1 (43)
Therefore: Tr
ε^21
= Tr(ε 1 ) = 1 (44) Conclusion: ε 1 is a pure state. For ε 2 = 12 | 0 → ↓ 0 | + 12 | 1 → ↓ 1 |: Calculate ε 22 :
ε 22 =
Therefore: Tr
ε (^22)
Conclusion: ε 2 is a mixed state. Part b) Answer: Matrix representation for ε 1 = |+→ ↓+|: First, compute |+→ ↓+|:
|+→ ↓+| =
In matrix form (computational basis {| 0 → , | 1 →}):
ε 1 =
Matrix representation for ε 2 = 12 | 0 → ↓ 0 | + 12 | 1 → ↓ 1 |:
ε 2 =
Physical explanation: ε 1 has o!-diagonal terms (the 1 entries in the upper-right and lower-left corners). These o!-diagonal elements represent quantum coherence—the system is in a coherent superposition of | 0 → and | 1 →. This coherence is characteristic of a pure state. ε 2 is diagonal with no o!-diagonal terms. This means there is no quantum co- herence—the system is equally in state | 0 → OR in state | 1 → with equal probability 1/ 2 each. This is a classical probabilistic mixture, not a quantum superposition, which is characteristic of a mixed state. The key di!erence: Quantum superposition (pure) vs. classical probability mixture (mixed).
Evaluating:
(I (^) A ↑ ↓ 1 | (^) B ) | 00 → ↓ 00 | (I (^) A ↑ | 1 → (^) B ) = 0 (since ↓ 1 | 0 → = 0) (64) (I (^) A ↑ ↓ 1 | (^) B ) | 01 → ↓ 01 | (I (^) A ↑ | 1 → (^) B ) = | 0 → (^) A ↓ 0 | (^) A · ↓ 1 | 1 → · ↓ 1 | 1 → = | 0 → ↓ 0 | (^) A (65) (I (^) A ↑ ↓ 1 | (^) B ) | 10 → ↓ 10 | (I (^) A ↑ | 1 → (^) B ) = 0 (since ↓ 1 | 0 → = 0) (66)
Therefore: (I (^) A ↑ ↓ 1 | (^) B )ε (^) AB (I (^) A ↑ | 1 → (^) B ) =
| 0 → ↓ 0 | A (67)
Step 3: Sum both contributions:
ε (^) A =
| 0 → ↓ 0 | A +
| 1 → ↓ 1 | A
| 0 → ↓ 0 | A =
| 0 → ↓ 0 | A +
| 1 → ↓ 1 | A (68)
Part b) Answer:
Tr(ε (^) A ) =
Tr(| 0 → ↓ 0 |) +
Tr(| 1 → ↓ 1 |) =
Interpretation: The trace equals 1, as required by the definition of a valid density operator. This confirms that ε (^) A describes a properly normalized quantum state where probabilities sum to 1. The fact that we start with a joint state (Tr(εAB ) = 1) and obtain a normalized reduced state (Tr(εA ) = 1) is a fundamental property of the partial trace operation. Part c) Answer: To determine if ε (^) A is pure or mixed, calculate Tr(ε^2 A ):
ε (^2) A =
Therefore: Tr
ε (^2) A
Conclusion: Since Tr(ε^2 A ) < 1, the state ε (^) A is mixed. This makes physical sense: the original joint state contains information about both qubits, but when we trace out (discard information about) qubit B, we lose information about the correlations between A and B. This lost information manifests as mixedness in the reduced state ε (^) A.
9.4 Question 4: Bell State and Entanglement (15 marks)
Question: The Bell state is defined as:
∣∣ $ +^
Part a) [5 marks] Write the joint density operator εAB = |$+^ →↓$ +^ | and compute its von Neumann entropy S(ε (^) AB ). What does this tell us about the purity of the joint state? Part b) [8 marks] Calculate the reduced density matrices ε (^) A = Tr (^) B (ε (^) AB ) and εB = Tr (^) A (ε (^) AB ) using the explicit definition of partial trace. Show that both reduced states are identical and characterize their degree of mixedness. Part c) [2 marks] Explain why the fact that the joint state is pure but the reduced states are mixed is a signature of entanglement.
Complete Solution
Part a) Answer: Joint density operator:
ε (^) AB =
$ +^
$ +^
Von Neumann entropy: Since ε (^) AB is a pure state, it has only one eigenvalue equal to 1 (and all others equal to 0). Therefore:
S(ε (^) AB ) = ⇐ Tr(εAB ln εAB ) = ⇐(1 · ln 1 + 0 · ln 0) = 0 (75) Physical interpretation: The zero entropy indicates that the joint system is in a pure state—it is completely determined with no quantum uncertainty. All information about the joint system is available; there is no missing information about the state of the composite system AB. Part b) Answer: Calculate εA = Tr (^) B (ε (^) AB ): Using basis {| 0 → (^) B , | 1 → (^) B }: For j = 0:
(I (^) A ↑ ↓ 0 | (^) B )ε (^) AB (I (^) A ↑ | 0 → (^) B ) (76)
=
(I A ↑ ↓ 0 | B )(| 00 → ↓ 00 | + | 00 → ↓ 11 | + | 11 → ↓ 00 | + | 11 → ↓ 11 |)(I A ↑ | 0 → B ) (77)
(| 0 → ↓ 0 | A · 1 · 1 + | 0 → ↓ 1 | A · 1 · 0 + | 1 → ↓ 0 | A · 0 · 1 + | 1 → ↓ 1 | A · 0 · 0) (78)
| 0 → ↓ 0 | A (79)
9.5 Question 5: Schmidt Decomposition (15 marks)
Question: Consider the two-qubit state:
|ω→ =
| 0 → A | 0 → B +
| 1 → A | 1 → B (89)
Part a) [5 marks] This state is already in Schmidt form. Identify:
- The Schmidt rank r
- The Schmidt coe#cients s (^) k
- The Schmidt bases {|ϱk → (^) A } and {|ςk → (^) B }
Part b) [8 marks] Compute the reduced density matrices ε (^) A and εB directly from the Schmidt decomposition. Verify that they are identical and show their spectral de- composition. Part c) [2 marks] What is the von Neumann entropy of εA? Interpret the result in terms of entanglement.
Complete Solution
Part a) Answer: The state is given as:
|ω→ =
| 0 → A | 0 → B +
| 1 → A | 1 → B (90)
This is already in Schmidt form: |ω→ =
k s^ k^ |ϱk^ →^ A ↑^ |ς^ k^ →^ B Therefore:
- Schmidt rank: r = 2 (two terms in the decomposition)
- Schmidt coe”cients: s 1 = ↑^12 , s 2 = ↑^12
- Schmidt bases:
- For subsystem A: {|ϱ 1 → (^) A , |ϱ 2 → (^) A } = {| 0 → (^) A , | 1 → (^) A }
- For subsystem B: {|ς 1 → (^) B , |ς 2 → (^) B } = {| 0 → (^) B , | 1 → (^) B } Verification: Check
k s^ (^2) k = 1: ( 1 ≃ 2
Part b) Answer: From the Schmidt decomposition, the reduced density matrices are:
εA = Tr (^) B (|ω→↓ω|) =
k
s (^2) k |ϱk →↓ϱ (^) k | (^) A =
| 0 →↓ 0 | A +
| 1 →↓ 1 | A (92)
ε (^) A =
| 0 → ↓ 0 | A +
| 1 → ↓ 1 | A (93)
Similarly:
εB = Tr (^) A (|ω→↓ω|) =
k
s (^2) k |ς (^) k →↓ς (^) k | (^) B =
| 0 → ↓ 0 | B +
| 1 → ↓ 1 | B (94)
Verification that they are identical: Both have the form 12 I (maximal mixture), confirming that both reduced states are identical (up to subsystem labeling). Spectral decomposition: Both ε (^) A and ε (^) B already are in spectral form: For εA : ε (^) A =
| 0 → ↓ 0 | A +
| 1 → ↓ 1 | A (95)
with eigenvalues ϖ 1 = ϖ 2 = 12 and orthonormal eigenvectors {| 0 → (^) A , | 1 → (^) A }. Key observation: Both εA and ε (^) B have the same spectrum {s^2 k } = { 1 / 2 , 1 / 2 }, confirming a fundamental property of Schmidt decomposition. Part c) Answer: Calculate the von Neumann entropy of ε (^) A :
S(ε (^) A ) = ⇐
k
s (^2) k ln
s (^2) k
ln(1/2) ⇐
ln(1/2) = ln 2 ⇒ 0 .693 nats (96)
Interpretation: For a bipartite pure state, the entropy S(ε (^) A ) = ⇐
k s^ (^2) k ln s (^2) k is called the entropy
of entanglement. It quantifies how entangled the two subsystems are:
- S(εA ) = 0 means no entanglement (unentangled/separable state, Schmidt rank =
- S(εA ) = ln 2 means maximal entanglement for a two-qubit system (Schmidt rank = 2) Here, S(ε (^) A ) = ln 2 indicates this is a maximally entangled state. The Bell state shows maximal entanglement between the two qubits.
