# Identity operator

1. Sep 28, 2011

### johng23

I am trying to follow a derivation in a book which is written without bra-ket notation, and presumably without the concept of state vectors. I can easily follow it if I may use the fact that $\sum_{n}|\varphi_{n}\rangle\langle\varphi_{n}|$ is the identity operator.

Analogously to the way I would prove that the above expression is the identity operator: I write $\psi=\sum_{n}c_{n}\varphi_{n}$ as the expansion of a wavefunction on the complete basis set $\varphi_{n}$. If I use the fact that $c_{m}=\int\varphi^{*}_{m}\psi d^{3}\textbf{r}$, I can write $\psi=\sum_{n}\varphi_{n}\int\varphi^{*}_{n}\psi d^{3}\textbf{r}$. The statement which I would like to prove is identical to this, If I replace $\psi$ on both sides by another expression $\hat{w}\varphi_{n}$. Thus I would like to extract $\psi$ from the RHS and equate the rest of the expression to identity. How can I do this? The operator $\hat{w}$ has no special relation to the basis functions $\varphi_{n}$.

2. Sep 28, 2011

### nonequilibrium

Hello,

To translate from braket notation to your notation:
$$\mid \phi_n \rangle \leftrightarrow \phi_n$$
$$\langle x \mid y \rangle \leftrightarrow \int x^* y \mathrm d^3 r$$

Note that in the braket notation one formally rewrites $\langle x \mid y \rangle$ as $\left( \langle x \mid \right) \left( \mid y \rangle \right)$ where the x-thing is seen as an operator working on the y-thing.

In this formalism, and translating, we get

$$\int \phi_m^* \psi \mathrm d^3 r \leftrightarrow \langle \phi_m \mid \psi \rangle = \left( \langle \phi_m \mid \right) \left( \mid \psi \rangle \right)$$

and thus:

$\boxed{ \mid \psi \rangle} \leftrightarrow \psi=\sum_{n}\varphi_{n}\int\varphi^{*}_{n}\psi d^{3}\textbf{r} \leftrightarrow \boxed{ \sum_n \mid \phi_n \rangle \langle \phi_m \mid \left( \mid \psi \rangle \right) }$

3. Sep 28, 2011

### johng23

Thanks for your answer. Actually though, I was more wondering whether it was possible to show that without introducing the concept of state vectors at all. The book doesn't use them so it seems there must be a way.

4. Sep 29, 2011

### nonequilibrium

Oh, I see, you didn't want to prove the bra-ket identity theorem, but rather wanted an analogous expression without the Dirac notation (such that my whole post was redundant), correct?

If I understand you correctly, I think this might be an answer:
yes you can,
just define a set of http://en.wikipedia.org/wiki/Linear_functional" [Broken] as $\omega_n: H \to \mathbb C: \psi \mapsto \int \phi_n(\textbf r)^* \psi(\textbf r) \mathrm d^3 \textbf r$ where H is the hilbert space that we're working in.

With this entity, you can see (using your calculations) that the operator $A : H \to H: \psi \mapsto \sum_n \phi_n \omega_n(\psi)$ is equal to the identity operator.

In shorthand, you can write $I = \sum_n \phi_n \omega_n$, where the definition of the operator is implicit. This is actually the same as what happens in the Dirac-notation, but less ambiguous.

Last edited by a moderator: May 5, 2017