# QM: Operator in momentum representation

## Homework Statement

Hi guys

As we have discussed earlier, we can represent some operator in an arbitrary basis by the use of the 1-operator:

$$T = \hat{1} T \hat{1} = \sum\limits_{\sigma_a,\sigma_b } {\left| {\psi _{\sigma_a} \left( {r_i } \right)} \right\rangle T_{\sigma_a\sigma_b \right\rangle \left\langle {\psi_{\sigma_b} \left( {r_i } \right)} \right|}$$

However, in my book they represent the kinetic energy operator in momentum space by the following (disregarding spin)

$$\left\langle {{\bf{k}}'} \right|T\left| {\bf{k}} \right\rangle \propto k^2 \delta _{{\bf{k}},{\bf{k}}'}.$$

I cannot seem to connect these two methods of representing operators in some basis. How can one realise that the book's way of transforming is the same as ours with 1-operators?

Best,
Niles.

Last edited:

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fzero
Homework Helper
Gold Member
The expression

$$\left\langle {{\bf{k}}'} \right|T\left| {\bf{k}} \right\rangle \propto k^2 \delta _{{\bf{k}},{\bf{k}}'}$$

is for the matrix elements of $$T$$. The corresponding operator could be written as

$$\hat{T} = \sum_k c k^2 |k\rangle\langle k|,$$

where $$c$$ is the proportionality constant (probably $$1/(2m)$$).

The expression

$$\left\langle {{\bf{k}}'} \right|T\left| {\bf{k}} \right\rangle \propto k^2 \delta _{{\bf{k}},{\bf{k}}'}$$

is for the matrix elements of $$T$$. The corresponding operator could be written as

$$\hat{T} = \sum_k c k^2 |k\rangle\langle k|,$$

where $$c$$ is the proportionality constant (probably $$1/(2m)$$).
Thanks, but how do we know what I have highlighted above? I can see that you have inserted the 1-operator.

fzero
$$T = \sum\limits_{\sigma_a,\sigma_b } {\left| {\psi _{\sigma_a} \left( {r_i } \right)} \right\rangle T_{\sigma_a\sigma_b \right\rangle \left\langle {\psi_{\sigma_b} \left( {r_i } \right)} \right|}$$