Two alternative expressions for the expectation value of energy are

\begin{align} \langle H\rangle = \langle \psi|H|\psi\rangle \end{align} which holds only for pure state, and \begin{align} \langle H\rangle = \text{Tr}(\rho H) \end{align} which holds for mixed states.

So that got me wondering, in the case where $\rho = |\psi\rangle\langle \psi|$, a pure state,

\begin{align} \langle H\rangle = \text{Tr}(|\psi\rangle\langle \psi|H) = \text{Tr}(\langle \psi|H|\psi\rangle) = \langle \psi|H|\psi\rangle \end{align}

I know that trace is cyclic w.r.t matrix product, i.e. $\text{Tr}(ABC) = \text{Tr}(BCA) = \text{Tr}(CAB)$, but does the above mean that it is also cyclic w.r.t tensor product? (Since $|\psi\rangle\langle \psi|$ is really just $|\psi\rangle \otimes \langle \psi|$). It intuitively feels wrong since you're basically ripping Hilbert space in half, but maybe even if it doesn't work in general there may be a restricted set of cases where it does work? What do people think?


1 Answer 1


$Tr(A\otimes B \otimes C)=Tr(A) Tr(B) Tr(C).$

So sure, it's also equal to $Tr(C\otimes A \otimes B)$, but cyclic isn't the property being used here.

Edit: Just caught myself: This only works when A, B, C are square so the trace is well-defined. For you, this is not the case, so I don't see how you can use the tensor product to explain this property. In this case you really have to use the fact that $|\psi \rangle \langle \psi |$ is also matrix multiplication:

If $|\psi \rangle$ is $n$ x $1$, then by ordinary matrix multiplication, $|\psi \rangle \langle \psi |$ is ($n$ x $1$)($1$ x $n$) = $n$ x $n$ matrix. I prefer this way of thinking about it, since usually tensor products separate different systems, but here the product is between two objects ($|\psi \rangle $ and $\langle \psi |$ ) from the same system, so it's a bit unorthodox - though mathematically not incorrect - to think of it as a tensor product.

Then we have

$$Tr(| \psi \rangle \langle \psi | H ) = Tr(\langle \psi | H | \psi \rangle )= \langle \psi | H | \psi \rangle$$

For the first equality, I used ordinary cyclicity of the trace: For matrix multiplication, the trace is cyclic for any product for which the matrix multiplication is still defined. Including for "$1$x$1$ matrices" like $\langle \psi | H |\psi \rangle$, whose trace is just themselves.


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