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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 $\psi_{i,j}$?

glS
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ZR-
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3 Answers3

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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.

Adam Zalcman
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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.

DaftWullie
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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.

joseph h
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