The trace of the outer product in Quantum Mechanics.

Question

I keep seeing the identity $$ \textbf{tr}(\left|\alpha\right>\left<\beta\right|)=\left<\beta\right|\left.\alpha\right> $$ which follows from the definition $$ \textbf{tr}A=\sum\limits_k\left<k\right|A\left|k\right>. $$

My confusion comes from when I try to derive the identity myself. I assume that $\left|\alpha\right>$ and $\left|\beta\right>$ are orthonormal, and I get $$ \textbf{tr}(\left|\alpha\right>\left<\beta\right|)=2\left<\beta\right|\left.\alpha\right>. $$ I think my issue is the assumption of orthonormality, but I don't know how that would change my result. Can someone show me how this derivation is done? Thank you.

my work

Answer 1

Start from: $$\textbf{tr}A=\sum\limits_k\left<k\right|A\left|k\right> $$ where the sum goes over an orthonormal basis.

Substitute $A = \left|\alpha\right>\left<\beta\right|$ . We get:$$\textbf{tr}(\left|\alpha\right>\left<\beta\right|)= \sum\limits_k\left<k\right| \;\left|\alpha\right>\left<\beta\right|\;\left|k\right> = \\ = \sum\limits_k\left<k\right| \left|\alpha\right>\;\left<\beta\right|\left|k\right> =\\= \sum\limits_k \left<\beta\right|\left|k\right> \;\left<k\right| \left|\alpha\right> =\\= \sum\limits_k \left<\beta\right|\;\left|k\right> \left<k\right| \;\left|\alpha\right> =\\= \left<\beta \right|\left(\sum\limits_k \left|k\right> \left<k\right| \right)\left|\alpha\right> = \\ = \left<\beta \right| I \left|\alpha\right> = \left<\beta\right|\left.\alpha\right> $$

Answer 2

There is way to establish the result in a simple manner using plainly "matrix notations" :

Let :

$$U:=\pmatrix{\alpha_1\\ \alpha_2 \\ \vdots \\ \alpha_n} \ \ \text{and} \ \ V:=\pmatrix{\beta_1\\ \beta_2 \\ \vdots \\ \beta_n}.$$

• First way :

$$UV^T=\pmatrix{\alpha_1\\ \alpha_2 \\ \vdots \\ \alpha_n}\pmatrix{\beta_1& \beta_2&\cdots&\beta_n}=\pmatrix{\alpha_1\beta_1 & \alpha_1\beta_2 & \cdots & \alpha_1 \beta_n\\ \alpha_2\beta_1 & \alpha_2\beta_2 & \cdots & \alpha_2 \beta_n\\ \cdots & \cdots &\cdots &\cdots \\ \alpha_n\beta_1 & \alpha_n\beta_2 & \cdots & \alpha_n \beta_n}\tag{1}$$

Therefore :

$$\operatorname{trace}UV^T=\alpha_1\beta_1+\alpha_2\beta_2+\cdots + \alpha_n\beta_n=U^TV$$

• Second way :

Let us recall property $$\operatorname{trace}(AB)=\operatorname{trace}(BA)\tag{2}$$

valid for any "compatible" matrices i.e., allowing both products $AB$ and $BA$ (otherwise said with resp. dimensions $n \times p$ and $p \times n$) for some $n$ and $p$, here with $p=1$.

$$\operatorname{trace}(UV^T)=\operatorname{trace}\underbrace{(V^TU)}_{\in \mathbb{R} \ }=\underbrace{V^TU}_{\text{dot product}}=U^TV,\tag{3}$$

the last equality being due to the fact that the dot product is commutative.