I'm studying QM from the book 'Quantum Mechanics. Concepts and Applications' by Zettili. There's an example which gives us two state vectors $$ | \psi \rangle = 9i \ | \phi_1 \rangle + 2 | \phi_2 \rangle $$ and $$ \langle \chi | = \frac{1}{\sqrt{2}} \big( i \langle \phi_1 | + \langle \phi_2 | \big) $$ where $ | \phi_1 \rangle $ and $| \phi_2 \rangle $ form an orthonormal and complete basis. Now, the author wants to compute the trace of the operator $ | \psi \rangle \langle \chi | $, which he does as follows: $$ \text{Tr}\big( | \psi \rangle \langle \chi | \big) = \text{Tr} \big( \langle \chi | \psi \rangle \big) = \langle \chi | \psi \rangle $$ He says this follows from the fact that $\text{Tr}(AB) = \text{Tr}(BA)$, for matrices. Still, I don't understand how he derives the second equality though. As far as I understand, the expression $ | \psi \rangle \langle \chi | $ is an operator, which corresponds to a square matrix. But by writing $\text{Tr}\big( | \psi \rangle \langle \chi | \big) = \text{Tr} \big( \langle \chi | \psi \rangle \big)$, it seems to me the author treats them like separate state vectors which correspond to square matrices (which they do not: they correspond to column vectors, for which trace is undefined), so that he can interchange the order.

And still, even if I grant him the second equality, how does he derive the third equality? Because this means he reduces the trace to a scalar product?

Some clarifications would be appreciated!

  • $\begingroup$ Let $| \psi \rangle$ be a vector with components $a_i$ and $\langle \chi |$ be the vector with components $b_i$. Then, the square matrix $ ( | \psi \rangle\langle \chi |)$ has components $( | \psi \rangle\langle \chi |)_{ij} = a_i b_j$. Then, it's trace is $a_i b_i$ which is also equal to the inner product of the two vectors, namely $\langle \chi | \psi \rangle$. $\endgroup$ – Prahar Mar 11 '16 at 21:27

Linear algebra result states that if $A$ is an $m\times n$ matrix and B is an $n\times m$ matrix, then $\text{Tr}(AB)=\text{Tr}(BA)$. The proof is elementary, using the definition of product of matrices. So in the case of vectors, let $A=|\psi⟩$ and $B=|\chi⟩$. Note that $A,B$ need not be square. It therefore follows that $\text{Tr}(|\psi⟩⟨\chi|)=\text{Tr}(⟨\chi|\psi⟩)$. The third equality follows from the fact that trace of a scalar is the same scalar, since a scalar is nothing but $1\times 1$ matrix, which you can verify since inner product of $1\times m$ and $m\times 1$ vectors is just $1\times 1$ matrix and the trace is just the entry itself.


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