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Why is it that, given two quantum states $|\psi_1\rangle$, $|\psi_2\rangle$, $$\mathrm{Tr}(|\psi_1\rangle\langle\psi_2|) = \langle\psi_2|\psi_1\rangle \quad $$ I went through the equation with the qubit space $\mathcal{H}=\{|0\rangle,|1\rangle\}$ and saw that the maths checks out, but I'm lacking any deeper understanding or intuition.

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2 Answers 2

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The calculation itself takes one line. Letting $|n\rangle$ be an arbitrary orthonormal basis, we have

$$\mathrm{Tr}\left(|\psi_1\rangle\langle \psi_2|\right)= \sum_n \langle n|\psi_1\rangle\langle \psi_2|n\rangle = \sum_n\langle \psi_2|n\rangle\langle n|\psi_1\rangle$$ $$ = \langle \psi_2|\left( \sum_n |n\rangle\langle n| \right)|\psi_1\rangle = \langle \psi_2|\psi_1\rangle$$

(Okay, two lines for readability.)

As far as intuition goes, the trace of an operator is equal to the sum of its eigenvalues. The eigenvalue equation for this operator is $$|\psi_1\rangle\langle \psi_2|\phi\rangle = \lambda |\phi\rangle \implies \langle \psi_2|\psi_1\rangle\langle \psi_2|\phi\rangle = \lambda \langle \psi_2|\phi\rangle$$ $$\implies \bigg(\langle \psi_2|\psi_1\rangle -\lambda\bigg) \langle \psi_2|\phi\rangle = 0$$

So either $\lambda = \langle \psi_2|\psi_1\rangle \implies |\phi\rangle = |\psi_1\rangle$ or $\langle \psi_2|\phi\rangle = 0$, in which case $\lambda = 0$. We can therefore conclude that if $|\psi_1\rangle$ and $|\psi_2\rangle$ are orthogonal, then all of the operator's eigenvalues (and therefore its trace) are zero; if they are not orthogonal, then the operator's single non-zero eigenvalue (and again, therefore its trace) is equal to $\langle \psi_2|\psi_1\rangle$.

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  • $\begingroup$ Example of something that takes one line but needs 3 lines for readability: Zagier's one-sentence proof $\endgroup$
    – mathlander
    Nov 25, 2022 at 17:51
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Traces are invariant under cyclical rotation. I.e. \begin{align} \mathrm{Tr}(ABCD) = \mathrm{Tr}(DABC) = \mathrm{Tr}(CDAB) = \mathrm{Tr}(BCDA) \end{align}

This holds for any number of matrices, and the individual matrices do not need to be square. Just their product must be square.

In the physics setting you should interpret $|\psi_1\rangle$ as a column vector (i.e. a tall, narrow matrix), and $\langle\psi_2|$ as a row vector (i.e. a flat, wide matrix). Furthermore $\langle\psi_2|\psi_1\rangle$ is just a number (i.e. a $1\times 1$ matrix), so we can write \begin{align} \mathrm{Tr}(|\psi_1\rangle\langle\psi_2|) = \mathrm{Tr}(\langle\psi_2|\psi_1\rangle)=\langle\psi_2|\psi_1\rangle \end{align}

It is actually a somewhat common trick in physics (for example when computing Feynman diagrams in QFT), to surround a single number with a $\mathrm{Trace}()$ and then cycle the terms around.

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