Chip said:
Maybe a picture will help. http://chipreuben.com/annihilation-of-shifted-Gaussian.jpg
I get zero, both with the shifted and the non-shifted. Can anyone tell why the nonshifted should yield the ground state eigenfunction multiplied up by a constant? It seems the answer has to do with expanding in Fourier mode and then converting back to the distance domain...but I'm not sure.
Okay, I’m going to work in units where \hbar = m = \omega = 1. In this units, the momentum operator is simply
p = \frac{i}{\sqrt{2}} \ (a^{\dagger} - a) . \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
Now, consider the translation operator, T(\epsilon) = \exp(- i \epsilon \ p), whose action on coordinate state is given by T(\epsilon) | x \rangle = | x + \epsilon \rangle , \ \ \ \ \ \ (2)
and define a state |\Psi_{\epsilon} \rangle by the action of T(\epsilon) on the ground state |\Psi_{0}\rangle:
<br />
|\Psi_{\epsilon} \rangle = T(\epsilon) \ | \Psi_{0}\rangle . \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)<br />
By multiplying (3) from the left with \langle x| and inserting the unit operator \int d \bar{x} |\bar{x}\rangle \langle \bar{x}| =1, we transfer (3) to a coordinate representation, i.e., equation between wave functions
<br />
\langle x | \Psi_{\epsilon}\rangle = \int d \bar{x} \ \langle x | T(\epsilon) | \bar{x} \rangle \langle \bar{x}| \Psi_{0}\rangle . \ \ \ \ (4)<br />
Using (2), we obtain
<br />
\Psi_{\epsilon}(x) = \int d \bar{x} \ \langle x | \bar{x} + \epsilon \rangle \ \Psi_{0}(\bar{x}) . \ \ \ \ (5)<br />
Now, change the integration variable \bar{x} = y - \epsilon, and use \langle x | y \rangle = \delta (x - y) to obtain
<br />
\Psi_{\epsilon}(x) = \int dy \ \langle x | y \rangle \ \Psi_{0}(y - \epsilon) = \Psi_{0}(x - \epsilon) . \ \ (6)<br />
So, \Psi_{\epsilon}(x) is the shifted ground state wavefunction \Psi_{0}(x - \epsilon).
Okay, now go back to (3) and let the operator a acts on it
<br />
a \ |\Psi_{\epsilon}\rangle = a \ e^{- i \epsilon \ p} | \Psi_{0} \rangle . \ \ \ \ \ \ (7)<br />
Since, a \ |\Psi_{0}\rangle = 0, we can replace the right-hand-side of (7) by the following commutator
<br />
a \ |\Psi_{\epsilon}\rangle = \left[a \ , \ e^{- i \epsilon \ p} \right] | \ \Psi_{0} \rangle . \ \ \ \ \ \ \ (8)<br />
If you now expand the exponential and use the relation
<br />
[a \ , -i \epsilon \ p ] = \frac{\epsilon}{\sqrt{2}}\ [a \ , a^{\dagger}] = \frac{\epsilon}{\sqrt{2}} ,<br />
you get
<br />
\left[ a \ , e^{- i \epsilon \ p} \right] = \frac{\epsilon}{\sqrt{2}} \ e^{- i \epsilon \ p} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ (9)<br />
Substituting (9) in (8), leads to
<br />
a \ |\Psi_{\epsilon}\rangle = \frac{\epsilon}{\sqrt{2}} \ e^{- i \epsilon \ p} \ | \Psi_{0} \rangle = \frac{\epsilon}{\sqrt{2}} \ |\Psi_{\epsilon} \rangle . \ \ \ \ \ \ (10)<br />
In wave function language, this becomes
<br />
a \ \Psi_{\epsilon}(x) = \frac{\epsilon}{\sqrt{2}} \ \Psi_{\epsilon} (x) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (11)<br />
And finally using (6), i.e., \Psi_{\epsilon}(x) = \Psi_{0}(x - \epsilon), we obtain
<br />
a \ \Psi_{0}( x - \epsilon ) = \frac{\epsilon}{\sqrt{2}} \ \Psi_{0}( x - \epsilon ) . \ \ \ \ \ \ \ \ \ \ (12)<br />
Notice that setting \epsilon = 0 gives you back
<br />
a \ \Psi_{0}(x) = 0 .<br />
