I'm trying to solve the following question:

"Prove that one way to compute $\mathrm Tr_B$ is to assume that someone has measured system $B$ in any orthonormal basis but does not tell you the measurement outcome." - "An Introduction to Quantum Computing" by Phillip Kaye.

Where $\mathrm{Tr}_B$, represents the partial trace with respect to some subsystem in say a bipartite system $H_A \otimes H_B$.

I have some reasoning behind it, but I don't see this as a proof. For example if we have say if we have $|\phi_1\rangle, |\phi_2\rangle \in H_A \otimes H_B$, $|\phi_1\rangle = |a_1\rangle \otimes |b_1\rangle$, $|\phi_2\rangle = |a_2\rangle \otimes |b_2\rangle$

So $|\phi_1\rangle\langle\phi_2| = |a_1\rangle\langle a_2| \otimes |b_1\rangle\langle b_2|$, and if someone were to measure the system $H_B$, then $|\phi_1\rangle\langle\phi_2|$ becomes the zero operator if $ |b_1\rangle \neq |b_2\rangle$. Since the $H_B$ component of the $|\phi_i\rangle$ has already collapsed only the operators on $H_B$ that don't change the state, i.e. where $|b_1\rangle = |b_2\rangle$ make sense. But since the info from $H_B$ was discarded we haven't learned anything about $H_A$, so the result is just $(\langle b_1 ||b_2\rangle)(|a_1\rangle\langle a_2|)$.

I kind of see how this becomes the partial trace, since $\mathrm{Tr}_B(|b_1\rangle \langle b_2|) = \langle b_1 ||b_2\rangle $.

I am looking for hints about how to extend this to a proof, or to fill in any gaps in my understanding.

  • 1
    $\begingroup$ Remark: It's also equivalent to just discarding. You can just as well not measure the qubit and say you did. Or you can throw it away and let someone else fish it out of the trash and measure it. $\endgroup$ Oct 29, 2019 at 2:23
  • $\begingroup$ related: quantumcomputing.stackexchange.com/q/4981/55 $\endgroup$
    – glS
    Feb 19, 2021 at 18:33

2 Answers 2


Consider a bipartite state $|\psi\rangle=\sum_{ij}\psi_{ij}|i\rangle\otimes|j\rangle$. In the following, I will work directly on the matrix elements of the objects involved.

Tracing out the second space amounts to the following mapping $$\psi_{ij}\rightarrow \rho_{ii'}\equiv\sum_j \psi_{ij}\bar\psi_{i'j}.\tag A$$ Now forget about the partial trace, and consider what happens when you measure the second system in the computational basis. You will get the $j$-th outcome with probability $p_j\equiv\sum_i |\psi_{ij}|^2$, and the post-measurement state will be some $\phi^{(j)}$ with indices $\phi_i^{(j)}=\frac{1}{\sqrt{p_j}}\psi_{ij}$.

So now we need to find a way to define an object which contains the information about all the possible post-measurement states with their associated probabilities. This is the "discarding measurement information" part. As it turns out, mapping ket vectors to density matrices makes this very easy, as probabilistic mixtures of states are simply weighted sums of density matrices. We therefore map each $\phi^{(j)}$ into the corresponding density matrix, $$\phi^{(j)}_i\rightarrow\rho^{(j)}_{ii'}\equiv \phi^{(j)}_i\bar\phi^{(j)}_{i'} =\frac{1}{p_j}\psi_{ij}\bar\psi_{i'j},$$ and now to "forget" about what state was measured, we simply sum these density matrices with the relative probabilities, which gives $$\sum_j p_j\rho^{(j)}_{ii'}=\sum_j \psi_{ij}\bar\psi_{i'j},$$ which, as you might notice, is the same as (A).

  • $\begingroup$ Ah, so we are basically "marginalizing" out one of the systems, from a probability standpoint. And so "discarding" is also converting "quantum uncertainty " to "classical uncertainty"? Since the two systems are no longer entangled, we are just accounting for the fact that the "person" who discarded the result doesn't know which state the second system collapsed to, right? $\endgroup$
    – dylan7
    Oct 16, 2019 at 23:07

Let me give you the structure for how you go about answering this question. Let $|\Psi\rangle$ be the pure state shared between Alice and Bob. Bob measures his system in an orthonormal basis $\{|\phi_i\rangle\}$. Thus, Bob gets an answer $i$ with probability $p_i$, and the overall system is left in the state $|\psi_i\rangle|\phi_i\rangle$.

Now, Alice does not know which result Bob got, all she knows is that with probability $p_i$ she has the state $|\psi_i\rangle$. Thus, what is the density matrix that she uses to describe her system?

Next, you want to explicitly calculate what $p_i$ and $|\psi_i\rangle$ are based on $|\Psi\rangle$ and $|\phi_i\rangle$. This gives you an explicit expression for Alice's density matrix.

Finally, you want to compare this to the partial trace. Remember that when you trace over one subsystem, you can use any orthonormal basis that you want, so we'll pick $\{|\phi_i\rangle\}$: $$ \text{Tr}_B(|\Psi\rangle\langle\Psi|)=\sum_i(I\otimes\langle\phi_i|)|\Psi\rangle\langle\Psi|(I\otimes|\phi_i\rangle) $$


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