3
$\begingroup$

Suppose there're two systems $A$ and $B$, if we're in a pure state $|\psi\rangle\in\mathbb{H}_a\otimes\mathbb{H}_b$. Let $\hat A$ be an operator acting on $\mathbb{H}_a$ and $|\psi\rangle=\sum_{i,j}\psi_{i,j}|a_i\rangle|b_j\rangle$. The expectation value of A with respect to the state $|\psi\rangle$ can be calculated as $$ \langle A\rangle_{\psi}= Tr_{\mathbb{H}_a\otimes\mathbb{H}_b}(A|\psi\rangle\langle\psi|)=\sum_{i,j}\langle a_i|\langle b_j|A\otimes\mathbb{1}_B|\psi\rangle\langle\psi|a_i\rangle|b_j\rangle=... $$

I'm very confused about where these two equalities coming from? In particular, where does the term $|\psi\rangle\langle\psi|$ in the sum come from? Can I understand its function as generating a $\psi$ matrix with different constants $\psi_{i,j}$?

$\endgroup$
0

3 Answers 3

3
$\begingroup$

You haven't really specified what you're happy with as a starting point. I suppose $$ \langle A\rangle_\psi=\langle\psi|A\otimes 1_B|\psi\rangle? $$

To see that the other expressions are equivalent, I usually find it more helpful to work backwards. Let's say we start from $$ \text{Tr}(A\otimes 1|\psi\rangle\langle \psi|), $$ well mathematically, we can take the trace by performing a sum over any orthonormal basis $\{|u_i\rangle\}$ in the Hilbert space, $$ \text{Tr}(B)=\sum_i\langle u_i|B|u_i\rangle. $$ One such basis is the separable basis $|a_i\rangle|b_j\rangle$ which immediately gives you the second equality.

Another choice of basis has $|u_1\rangle=|\psi\rangle$, and all other states orthogonal. In this case $$ \langle u_i|A\otimes 1|\psi\rangle\langle\psi|u_i\rangle=\delta_{i,1}\langle\psi|A\otimes 1|\psi\rangle, $$ as required to get the first equality, using the definition I gave above.

$\endgroup$
3
$\begingroup$

It is a variant of the formula for the average $\langle A\rangle_\rho$ of an operator $A$ measured on a state $\rho$

$$ \langle A\rangle_\rho = \mathrm{tr}(A\rho).\tag1 $$

In this case we are measuring an operator acting on the first subsystem, so $\rho = \mathrm{tr}_B(|\psi\rangle\langle\psi|)$. Substituting into $(1)$, we get

$$ \begin{align} \langle A\rangle_\psi &= \mathrm{tr}(A \,\mathrm{tr}_B(|\psi\rangle\langle\psi|)) \\ &= \mathrm{tr}(\mathrm{tr}_B(A\otimes I |\psi\rangle\langle\psi|)) \\ &= \mathrm{tr}(A\otimes I |\psi\rangle\langle\psi|) \\ &= \mathrm{tr}(A|\psi\rangle\langle\psi|) \end{align} $$

where the first equality is the substitution, the second follows from the properties of partial trace, the third from the fact that tracing over subsystem $B$ followed by tracing over subsystem $A$ is equivalent to tracing over both of the subsystems and finally the fourth is a shortcut notation that allows to make identity implicit when composing operators defined on subsystems of a composite system.


The second equality follows from the fact that

$$ \mathrm{tr} A = \sum_k \langle k|A|k\rangle $$

where $|k\rangle$ is any orthonormal basis. This can be proven by first choosing the eigenbasis of $A$ and then appealing to basis-independence of the trace.

$\endgroup$
2
$\begingroup$

That is the definition of the expectation value of an operator $\hat A$ using the trace

$$\langle \hat A \rangle = Tr (\rho \hat A)$$

where the density matrix is given by

$$\rho = \mid \psi \rangle \langle \psi \mid $$

The density matrix is ubiquitous in quantum mechanics.

$\endgroup$

Your Answer

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

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