# Trace of operator defined by two state vectors in Quantum Mechanics

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!

• 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 $\langle \chi | \psi \rangle$. – 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.