How to prove that ${\rm tr}(A|\psi\rangle\langle\psi|)=\langle\psi| A|\psi\rangle$?

Question

How can one prove that $tr(A\mid\psi\rangle\langle\psi\mid)=\langle\psi\mid A\mid\psi\rangle$? In Nielsen/Chuang they mention this is due to Gram-Schmidt decomposition but I can’t understand how.

Answer 1

You can proceed in the following way: $\text{Tr}(A|\psi\rangle\langle\psi|) = \sum_i \langle i|A|\psi\rangle\langle\psi|i\rangle = \sum_i \langle\psi|i\rangle\langle i|a|\psi\rangle = \langle \psi|(\sum_i |i\rangle\langle i|)A|\psi\rangle = \langle\psi|A|\psi\rangle$

Answer 2

There's a couple of ways you can do this. Let $\{|u_i\rangle\}$ be any orthonormal basis. Then the definition of the trace is simply $$ \text{Tr}(A|\psi\rangle\langle\psi|)=\sum_i\langle u_i|A|\psi\rangle\langle\psi|u_i\rangle. $$ Since $\langle u_i|A|\psi\rangle\langle\psi|u_i\rangle$ is just two numbers multiplied together, I can change the order of multiplication to $\langle\psi|u_i\rangle\langle u_i|A|\psi\rangle$, so that the trace is $$ \text{Tr}(A|\psi\rangle\langle\psi|)=\sum_i\langle\psi|u_i\rangle\langle u_i|A|\psi\rangle=\langle\psi|A|\psi\rangle. $$ by using the completeness relation.

One step beyond this I personally find to be particularly elegant. If I can choose any orthonormal basis, why don't I select the basis such that $|u_1\rangle=|\psi\rangle$. The other $|u_i\rangle$ can be absolutely anything, it really doesn't matter (yes, you could use Gram-Schmidt to construct them, but you never need to construct them). Then, you have $$ \text{Tr}(A|\psi\rangle\langle\psi|)=\sum_i\langle u_i|A|\psi\rangle\langle\psi|u_i\rangle=\langle u_1|A|\psi\rangle\langle\psi|u_1\rangle+\sum_{i\neq 1}\langle u_i|A|\psi\rangle\langle\psi|u_i\rangle. $$ All the terms in the final sum are 0 because the $|u_i\rangle$ are orthogonal to $|\psi\rangle=|u_1\rangle$. This just leaves the first term $$ \text{Tr}(A|\psi\rangle\langle\psi|)=\langle u_1|A|\psi\rangle\langle\psi|u_1\rangle=\langle\psi|A|\psi\rangle. $$

Answer 3

You can use the cyclic property of the trace, ${\rm Tr}(XY) = {\rm Tr}(YX)$.

Another way is to note that both sides are linear over $A$. Thus it's enough to prove it for $A = E_{ij} = |i\rangle\langle j|$.