I remember from QM that the completeness relation says

$$ \sum_{n=1}^\infty |e_n\rangle \langle e_n | = I$$

so that $\langle x\mid y\rangle =\sum_{n=1}^\infty \langle x\mid e_n\rangle \langle e_n \mid y\rangle$.

I was recently trying to prove a result on Trace operators and one calculation was

$$\sum_{k=1}^n \langle A g_k , h_k \rangle = \sum_{k=1}^n \operatorname{tr}(A(g_k\otimes h_k))$$ Where apparently if $f\in X^*$ and $y\in Y$ we define $y\otimes f:X\to Y$ by $(y\otimes f)(x) = f(x)y$; my attempt at this:

$$\sum_{k=1}^n \operatorname{tr}(A(g_k\otimes h_k)) = \sum_{k=1}^n\sum_{i=1}^\infty \langle A(g_k \otimes h_k) e_i, e_i\rangle$$ $$= \sum_{k=1}^n\sum_{i=1}^\infty \langle A(\langle e_i,h_k\rangle g_k), e_i\rangle = \sum_{k=1}^n\sum_{i=1}^\infty \langle Ag_k,e_i\rangle\langle e_i,h_k\rangle$$

So using my naive approach from quantum mechanics I just conclude that the last term is $\sum_{k=1}^n \langle A g_k , h_k \rangle $.

However, I feel uneasy about this because $\langle \cdot , \cdot \rangle$ is an inner product, while $\langle \cdot \mid \cdot \rangle$ is the bra-ket notation... whatever that means.

1) Can I apply the completeness relation to make my conclusion?

2) Is there a canonical relation between $\langle \cdot, \cdot \rangle$ and $\langle \cdot \mid \cdot \rangle$?

3) How can I prove the completeness relation (I believe it's an axiom in QM, but I reckon its equivalence (in functional analysis (if it exists)) is a theorem).

  • $\begingroup$ If I may, I would suggest thinking about this in the finite-dimensional setting first, using matrices to guide you. $\endgroup$ Jul 22, 2016 at 0:42
  • $\begingroup$ How would you possibly split up the inner product to get something like first centred equation? $\endgroup$
    – Squirtle
    Jul 22, 2016 at 0:56
  • $\begingroup$ @squirtle Are you familiar with Dirac's notation in terms of "bras" and "kets?" The first centered equation is simply an operator on a bra and a ket. $\endgroup$
    – Mark Viola
    Jul 22, 2016 at 1:02
  • $\begingroup$ I've come across the bra ket notation but never really understood it; so, for example, I don't know what you mean by "the ... is simply an operator on a bra and a ket." $\endgroup$
    – Squirtle
    Jul 22, 2016 at 1:15
  • $\begingroup$ 1) Yes, your derivation is perfectly fine. 2) Both are inner products. Only that in physics, the inner product is usually assumed to be linear in the second variable. 3) The completeness relation is equivalent to \begin{align*} x & =\sum_{n}\left\langle e_{n},x\right\rangle e_{n}=\sum_{n}\left|e_{n}\right\rangle \left\langle e_{n},x\right\rangle, \end{align*} valid for all x in the Hilbert space. $\endgroup$
    – James
    Jul 22, 2016 at 2:09

1 Answer 1


Theorem: Let $H$ be a real or complex Hilbert space, and let $\{e_{\alpha}\}_{\alpha\in\Lambda}$ be an orthonormal subset of $H$. The following are equivalent:

1. $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ is a complete orthonormal set, meaning that the only $x\in H$ that is orthogonal to every $e_{\alpha}$ is $x=0$.

2. Parseval's identity $\|x\|^2=\sum_{\alpha\in\Lambda}|\langle x,e_{\alpha}\rangle|^2$ holds for all $x \in H$.

3. The identity $\langle x,y\rangle = \sum_{\alpha\in\Lambda}\langle x,e_{\alpha}\rangle\langle e_{\alpha},y\rangle$ holds for all $x,y\in H$.

4. The subspace consisting of all finite linear combinations of the $e_{\alpha}$ is dense in $H$.

The conventions in Quantum provided for ket vectors $|x\rangle$. The bras are linear functionals. On a Hilbert space there is a canonical map $x \mapsto x^*$ from vectors to linear functionals given by $x^*(y)=\langle y,x\rangle$. In this way, you may treat $\langle y | x\rangle$ in the same ways as $\langle x,y\rangle$ by use of the canonical map.

  • $\begingroup$ What does "finite limit combinations" mean? $\endgroup$
    – Clclstdnt
    Jul 28, 2016 at 4:00
  • $\begingroup$ @Clclstdnt : It means a type. I changed that to "linear" now. $\endgroup$ Jul 28, 2016 at 11:51

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