I'm reading the book Quantum computation and quantum information by Mike & Ike and I'm stuck at 2.60/2.61. There, the author says that, given the operator $A|ψ⟩⟨ψ|$, its trace is:

$${\rm tr}(A|\psi\rangle\langle\psi|) = \sum\limits_i\langle i|A|\psi\rangle\langle\psi|i\rangle$$

Why would that be true? Why can we rearrange the bras and kets like that?

  • $\begingroup$ I worked backwards from equation 2.61 as follows but I'm concerned that my argument is circular so I will post as a comment: $$\langle\psi|A|\psi\rangle =\sum_{i'}\sum_{i''}\langle\psi| i'\rangle\langle i'| A| i''\rangle\langle i''|\psi\rangle $$ $$=\sum_{i'}\sum_{i''}\langle i'| A| i''\rangle \langle i''|\psi\rangle \langle\psi|i'\rangle =\sum_{i'}\langle i'|A|\psi\rangle\langle\psi|i'\rangle =tr(A|\psi\rangle\langle\psi|)$$ $\endgroup$
    – Julien
    Mar 18 '14 at 21:35
  • 9
    $\begingroup$ An entry of a matrix $M$ in Dirac notation is obtained (given a basis $\{|i\rangle\}$) via $M_{ij}=\langle i|M|j\rangle$. The trace is the sum of the diagonal entries, i.e. $\operatorname{tr}(M)=\sum_i M_{ii}$ and that's it... $\endgroup$
    – Martin
    Mar 18 '14 at 23:50
  1. Let $\{|i\rangle\}$ be an orthonormal basis for the Hilbert space of the system. Then the trace of an operator $O$ is given by (See the Addendum below) \begin{align} \mathrm {tr}(O) = \sum_i \langle i|O|i\rangle \end{align}

  2. For a given state $|\psi\rangle$, we define an operator $P_\psi$ by \begin{align} P_\psi|\phi\rangle = \langle\psi|\phi\rangle|\psi\rangle. \end{align} As a shorthand, we usually write $P_\psi = |\psi\rangle\langle\psi|$.

  3. Using steps 1 and 2, we compute: \begin{align} \mathrm{tr}(A|\psi\rangle\langle\psi|) &= \mathrm{tr}(A P_\psi) \\ &= \sum_i \langle i|AP_\psi|i\rangle\\ &= \sum_i \langle i|A (\langle\psi|i\rangle|\psi\rangle)\\ &= \sum_i \langle i|A|\psi\rangle\langle\psi|i\rangle \end{align} which is the desired result.

Addendum. (Formula for the trace)

For simplicity, I'll restrict the discussion to finite-dimensional vector spaces. Recall that if $O$ is a linear operator on a vector space $V$, and if $ \{|i\rangle\}$ is a basis for $V$, then the matrix elements $O_{ij}$ of $O$ with respect to this basis are defined by it's action on this basis as follows: \begin{align} O|i\rangle = \sum_jO_{ji}|j\rangle. \tag{$\star$} \end{align} The trace of the linear operator with respect to this basis is then defined as the sum of its diagonal entries; \begin{align} \mathrm{tr}(O) = \sum_i O_{ii}. \tag{$\star\star$} \end{align} Now it turns out that the trace is a basis-independent number, so we can simply refer to the trace of the the linear operator; it's just the trace with respect to any chosen basis.

Now, suppose that $V$ is equipped with an inner product, like in the case of Hilbert spaces, and let $\{|i\rangle\}$ be an orthonormal basis for $V$, then we can take the inner product of both sides of $(\star)$ with respect to an element $|k\rangle$ of the basis to obtain \begin{align} \langle k|O|i\rangle = \sum_j \langle k|O_{ji}|j\rangle = \sum_j O_{ji}\langle k|j\rangle = \sum_jO_{ji}\delta_{jk} = O_{ki} \end{align} In other words, $\langle k|O|j\rangle$ gives precisely the matrix element $O_{kj}$ of $O$ in the given basis. In particular, the diagonal entries are given by $\langle i|O|i\rangle$. Plugging this into $(\star\star)$, we get \begin{align} \mathrm{tr} (O) = \sum_i \langle i|O|i\rangle \end{align} as desired.

  • $\begingroup$ Right, my question is mainly why is the trace of an operator given by that thing you said, $\sum_i \langle i|O|i\rangle$. $\endgroup$ Mar 18 '14 at 21:58
  • 1
    $\begingroup$ @PedroCarvalho Ah ok. See the addendum I just wrote. $\endgroup$ Mar 19 '14 at 1:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.