Suppose the quantum system has state $$a|0_A\rangle|0_B\rangle + b|1_A\rangle|1_B\rangle.$$ I want to prove that system $A$ has reduced density matrix of $a^2|0_A\rangle\langle 0_A| + b^2 |1_A\rangle\langle 1_A|$ when $0_A\rangle$ and $|1_A\rangle$ are orthogonal (and assume that $0_B\rangle$ and $1_B\rangle$ are orthogonal), but I am having a hard time mathematizing calculations.

So how is the trace over $B$ mathematically done?

  • $\begingroup$ What did you do so far? $\endgroup$
    – Gonenc
    Commented Jan 29, 2018 at 18:58
  • $\begingroup$ Writing down a tensor product and trying to trace out. But I am confused on tracing out. What would it mean math-wise to trace out over subsystem $B$? That's the question I have. $\endgroup$ Commented Jan 29, 2018 at 19:41

2 Answers 2


The partial trace $\mathrm{Tr}_B$ is defined as:

  • a linear map $\mathrm{Tr}_B:\mathcal{L}(\mathcal H_A\otimes \mathcal H_B) \to \mathcal{L}(\mathcal H_A)$ that goes from the space $\mathcal{L}(\mathcal H_A\otimes \mathcal H_B)$ of linear operators on $\mathcal H_A\otimes \mathcal H_B$ to the space of linear operators on $\mathcal H_A$, such that
  • when it is given a tensor-product operator $O_A\otimes O_B$, it returns the left half and takes the trace of the right half, i.e. $\mathrm{Tr}_B(O_A\otimes O_B) = O_A\,\mathrm{Tr}(O_B).$

Note that the second half is not enough to specify what happens to arbitrary operators, but since you can decompose any arbitrary (nice-enough) operator on $\mathcal H_A\otimes \mathcal H_B$ as a sum of tensor-product terms, then the requirement that the partial trace be linear completely fixes its behaviour.

To calculate its action on an operator like \begin{align} \rho & = |\psi\rangle\langle\psi| \\ & = |a|^2 |0_A 0_B\rangle \langle 0_A 0_B| + a b^* |0_A 0_B\rangle \langle 1_A 1_B| \\ & \qquad + a^* b |1_A 1_B\rangle \langle 0_A 0_B| + |b|^2 |1_A 1_B\rangle \langle 1_A 1_B|, \\ & = |a|^2 |0_A \rangle \langle 0_A|\otimes |0_B\rangle \langle 0_B| + a b^* |0_A \rangle \langle 1_A|\otimes |0_B\rangle \langle 1_B| \\ & \qquad + a^* b |1_A \rangle \langle 0_A|\otimes |1_B\rangle \langle 0_B| + |b|^2 |1_A \rangle \langle 1_A|\otimes |1_B\rangle \langle 1_B|, \end{align} you have to apply those two properties sequentially: first you break up the sum, \begin{align} \mathrm{Tr}_B(\rho) & = |a|^2 \,\mathrm{Tr}_B(|0_A \rangle \langle 0_A|\otimes |0_B\rangle \langle 0_B|) + a b^* \, \mathrm{Tr}_B(|0_A \rangle \langle 1_A|\otimes |0_B\rangle \langle 1_B| ) \\ & \qquad + a^* b \,\mathrm{Tr}_B(|1_A \rangle \langle 0_A|\otimes |1_B\rangle \langle 0_B| ) + |b|^2 \, \mathrm{Tr}_B(|1_A \rangle \langle 1_A|\otimes |1_B\rangle \langle 1_B|), \end{align} then you pass the traces to the right-hand side, \begin{align} \mathrm{Tr}_B(\rho) & = |a|^2 |0_A \rangle \langle 0_A|\,\mathrm{Tr}_B( |0_B\rangle \langle 0_B|) + a b^* |0_A \rangle \langle 1_A|\, \mathrm{Tr}_B(|0_B\rangle \langle 1_B| ) \\ & \qquad + a^* b |1_A \rangle \langle 0_A|\,\mathrm{Tr}_B( |1_B\rangle \langle 0_B| ) + |b|^2 |1_A \rangle \langle 1_A| \,\mathrm{Tr}_B( |1_B\rangle \langle 1_B|), \end{align} and then you calculate the traces: \begin{align} \mathrm{Tr}_B(\rho) & = |a|^2 |0_A \rangle \langle 0_A| \times 1 + a b^* |0_A \rangle \langle 1_A| \times 0 \\ & \qquad + a^* b |1_A \rangle \langle 0_A| \times 0 + |b|^2 |1_A \rangle \langle 1_A| \times 1 \\ & = |a|^2 |0_A \rangle \langle 0_A| + |b|^2 |1_A \rangle \langle 1_A|. \end{align} Easy!

As some additional practice, try calculating the partial trace of $\rho=|\psi⟩⟨\psi|$ as you go from an entangled state to a separable one, by taking, say, $$ \psi = a|0_A\rangle|0_B\rangle + b|1_A\rangle|1_B\rangle +\sin(\theta)\sqrt{ab}(|0_A\rangle|1_B\rangle + b|1_A\rangle|0_B\rangle), $$ which simplifies to your state at $\theta=0$ and to the product state $$(\sqrt{a}|0_A⟩+\sqrt{b}|1_A⟩) \otimes (\sqrt{a}|0_B⟩+\sqrt{b}|1_B⟩)$$ at $\theta=\pi/2$: you should see that as the parameter $\theta$ increases, the relevance of the off-diagonal elements of $\mathrm{Tr}_B(\rho)$ increases as well.


Let's assume that the states $|0\rangle,|1\rangle$ are normalized. Let's also assume $|a|^2+|b|^2 = 1$ so that $|\psi\rangle = a |0_A 0_B\rangle+ b |1_A 1_B\rangle$ is normalized (this is all done so that $\rho$ is properly normalized).

We have then $\rho = |\psi\rangle\langle\psi|= |a|^2 |0_A 0_B\rangle \langle 0_A 0_B| + a b^* |0_A 0_B\rangle \langle 1_A 1_B| + a^* b |1_A 1_B\rangle \langle 0_A 0_B| + |b|^2 |1_A 1_B\rangle \langle 1_A 1_B|$.

Taking the trace over $B$ we get $\rho_A = \langle 0_B| \rho|0_B\rangle + \langle 1_B| \rho|1_B\rangle = |a|^2 |0_A\rangle \langle 0_A| + |b|^2 |1_A\rangle \langle 1_A|$.

  • $\begingroup$ The question is how is trace being done mathematically? $\endgroup$ Commented Jan 29, 2018 at 21:12
  • $\begingroup$ How formal do you want your definition? Check the answers in this link. $\endgroup$
    – secavara
    Commented Jan 29, 2018 at 22:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.