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Expectation values with annihilation/creation operators

  • #1

Homework Statement



Calculate [itex]<i(\hat{a} - \hat{a^{t}})>[/itex]


Homework Equations



[itex]|\psi > = e^{-\alpha ^{2}/2} \sum \frac{(\alpha e^{i\phi })^n}{\sqrt{n!}} |n>[/itex]

[itex] \hat{a}|n> = \sqrt{n}|n-1>[/itex]

I derived:
[itex] \hat{a}|\psi> = (\alpha e^{i\phi})^{-1}|\psi>[/itex]

The Attempt at a Solution



[itex]<i(\hat{a} - \hat{a^{t}})> = <\psi|i\hat{a}-i\hat{a^{t}}|\psi>
[/itex]

[itex]
<\psi|i\hat{a}-i\hat{a^{t}}|\psi> = <\psi|i\hat{a}|\psi> - <\psi|i\hat{a^{t}}|\psi>
[/itex]

[itex]
<\psi|i\hat{a}|\psi> - <\psi|i\hat{a^{t}}|\psi> = <\psi|i\hat{a}|\psi> - <-i\hat{a}\psi|\psi>
[/itex]

[itex]
i(\alpha e^{i\phi})^{-1}<\psi|\psi> + i(\alpha e^{i\phi})^{-1}<\psi|\psi>
[/itex]

assuming [itex] \psi [/itex] is normalized,

[itex]
<\psi|\psi> = 1
[/itex]

[itex]
<i(\hat{a} - \hat{a^{t}})> = 2i(\alpha e^{i\phi})^{-1}
[/itex]

Now, I think I did this correctly.. What I don't understand is the significance of
[itex]
<i(\hat{a} - \hat{a^{t}})>
[/itex]

Normally with expectation values, you can usually tell if your result is at least reasonable.. I don't understand what this expectation value is telling me, so I can't tell if my result is reasonable. =/

Any help would be much appreciated!!
 

Answers and Replies

  • #2
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
It would probably help to write ##\hat{a}, \hat{a}^\dagger## in terms of the position and momentum operators.
 

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