Annihilation of shifted gaussian vs. nonshifted gaussian

[Chip]

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.

 

[vanhees71]

The most simple thing is to just check the equation. You've given the ground-state wave function (a Gaussian) and then you can simply apply the annihilation operator and check the given relation. That shows that the "shifted ground state" is a coherent state.

 

[samalkhaiat]

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 $|\Psi_{0}\rangle$:
$$|\Psi_{\epsilon} \rangle = T(\epsilon) \ | \Psi_{0}\rangle . \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$
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
$$\langle x | \Psi_{\epsilon}\rangle = \int d \bar{x} \ \langle x | T(\epsilon) | \bar{x} \rangle \langle \bar{x}| \Psi_{0}\rangle . \ \ \ \ (4)$$
Using (2), we obtain
$$\Psi_{\epsilon}(x) = \int d \bar{x} \ \langle x | \bar{x} + \epsilon \rangle \ \Psi_{0}(\bar{x}) . \ \ \ \ (5)$$
Now, change the integration variable $\bar{x} = y - \epsilon$, and use $\langle x | y \rangle = \delta (x - y)$ to obtain
$$\Psi_{\epsilon}(x) = \int dy \ \langle x | y \rangle \ \Psi_{0}(y - \epsilon) = \Psi_{0}(x - \epsilon) . \ \ (6)$$
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
$$a \ |\Psi_{\epsilon}\rangle = a \ e^{- i \epsilon \ p} | \Psi_{0} \rangle . \ \ \ \ \ \ (7)$$
Since, $a \ |\Psi_{0}\rangle = 0$, we can replace the right-hand-side of (7) by the following commutator
$$a \ |\Psi_{\epsilon}\rangle = \left[a \ , \ e^{- i \epsilon \ p} \right] | \ \Psi_{0} \rangle . \ \ \ \ \ \ \ (8)$$
If you now expand the exponential and use the relation
$$[a \ , -i \epsilon \ p ] = \frac{\epsilon}{\sqrt{2}}\ [a \ , a^{\dagger}] = \frac{\epsilon}{\sqrt{2}} ,$$
you get
$$\left[ a \ , e^{- i \epsilon \ p} \right] = \frac{\epsilon}{\sqrt{2}} \ e^{- i \epsilon \ p} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ (9)$$
Substituting (9) in (8), leads to
$$a \ |\Psi_{\epsilon}\rangle = \frac{\epsilon}{\sqrt{2}} \ e^{- i \epsilon \ p} \ | \Psi_{0} \rangle = \frac{\epsilon}{\sqrt{2}} \ |\Psi_{\epsilon} \rangle . \ \ \ \ \ \ (10)$$
In wave function language, this becomes
$$a \ \Psi_{\epsilon}(x) = \frac{\epsilon}{\sqrt{2}} \ \Psi_{\epsilon} (x) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (11)$$
And finally using (6), i.e., $\Psi_{\epsilon}(x) = \Psi_{0}(x - \epsilon)$, we obtain
$$a \ \Psi_{0}( x - \epsilon ) = \frac{\epsilon}{\sqrt{2}} \ \Psi_{0}( x - \epsilon ) . \ \ \ \ \ \ \ \ \ \ (12)$$
Notice that setting $\epsilon = 0$ gives you back
$$a \ \Psi_{0}(x) = 0 .$$