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QM: Operator in momentum representation

  • Thread starter Niles
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  • #1
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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:

[tex]
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|}
[/tex]

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

[tex]
\left\langle {{\bf{k}}'} \right|T\left| {\bf{k}} \right\rangle \propto k^2 \delta _{{\bf{k}},{\bf{k}}'}.
[/tex]

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:

Answers and Replies

  • #2
fzero
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The expression

[tex]
\left\langle {{\bf{k}}'} \right|T\left| {\bf{k}} \right\rangle \propto k^2 \delta _{{\bf{k}},{\bf{k}}'}
[/tex]

is for the matrix elements of [tex]T[/tex]. The corresponding operator could be written as

[tex] \hat{T} = \sum_k c k^2 |k\rangle\langle k|,[/tex]

where [tex]c[/tex] is the proportionality constant (probably [tex]1/(2m)[/tex]).
 
  • #3
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The expression

[tex]
\left\langle {{\bf{k}}'} \right|T\left| {\bf{k}} \right\rangle \propto k^2 \delta _{{\bf{k}},{\bf{k}}'}
[/tex]

is for the matrix elements of [tex]T[/tex]. The corresponding operator could be written as

[tex] \hat{T} = \sum_k c k^2 |k\rangle\langle k|,[/tex]

where [tex]c[/tex] is the proportionality constant (probably [tex]1/(2m)[/tex]).
Thanks, but how do we know what I have highlighted above? I can see that you have inserted the 1-operator.
 
  • #4
fzero
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Homework Helper
Gold Member
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All I did was substitute the matrix elements into the corresponding version of the equation


[tex]

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|}

[/tex]
 
  • #5
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Thanks!
 

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