# Is Trace cyclic with respect to tensor product?

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?

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

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.

• PS: A very useful wikipedia page to land on is the Kronecker product, which is the matrix representation of the tensor product. en.wikipedia.org/wiki/Kronecker_product Dec 8, 2019 at 11:03