Some questions on coherent states and corresponding Hilbert spaces. Mathematical formalism.

Question

I have a few questions related to coherent states. I'm trying to understand the topic using this source.

1. Does the set of all coherent states form a Hilbert space? Or it forms the dense subset only? How is this space described?
2. Let $x\in \mathbb{R}$, then the corresponding coherent state $$|x\rangle=\exp\{-x^2/2\}\sum^\infty_{n=0}\frac{x^n}{\sqrt{n!}}|n\rangle.$$ Do I understand correctly that using a inner product $\langle\cdot|\cdot\rangle$ we can calculate the value of a coherent state (as a function) at a point?

For example, $\langle x | \alpha \rangle=\exp\{-x^2/2\}\sum^\infty_{n=0}\frac{x^n}{\sqrt{n!}}\,\phi_n(x)=|\alpha\rangle(x)$, where $\phi_n=|n\rangle$.

Answer 1

1. Yes, coherent states are an overcomplete basis for the Hilbert space a quantum oscillator acts on. Its energy eigenstates $|n\rangle$ are a complete orthonormal basis, and the coherent states are complete, $$\frac{1}{\pi} \int |\alpha\rangle\langle\alpha| d^2\alpha = I \qquad d^2\alpha \equiv d\Re(\alpha) \, d\Im(\alpha) ~,$$ but they are not orthogonal, $$\langle\beta|\alpha\rangle=e^{-{1\over2}(|\beta|^2+|\alpha|^2-2\beta^*\alpha)}\neq\delta(\alpha-\beta),$$ as eigenvectors of the non-self-adjoint annihilation operator $a$. Thus, if your oscillator is in the quantum state $|\alpha \rangle $, it is also with nonzero probability in the other quantum state $|\beta \rangle$ (but the farther apart the states are situated in phase space, the lower the probability is).

2. No, a dot product of two kets is a scalar, a function of x here, a wavefunction (Schroedinger's famed minimal wavepacket), and not a ket (vector) the odd way you wrote things. The oscillator wavefunction of energy n (+1/2) is $\langle x| n\rangle =\phi_n(x)$, a Hermite function. You may ascertain dotting $\langle m|$ onto your odd $|x\rangle$ does not yield such a function, but, instead, $x^m \exp (-x^2/2)$. Consequently, instead, for the nonstandard/eccentric $|x\rangle$ you write down, $\langle x |\alpha \rangle=\alpha(x)=\exp (-(x-\alpha)^2/2) $.

3. I can see the next question coming, so I'm answering it preemptively, right away. Normally, $$ \hat x |x\rangle=x|x\rangle= \frac{1}{\sqrt 2} (a + a^\dagger)|x\rangle , $$ and that is how it is defined in your linked notes. Instead you define your odd state as $$ a|x\rangle= x|x\rangle , $$ dangerously, and this is what I'm providing the "wavefunction" for, as per your expression. Now, intriguingly, for the conventional $|x\rangle$ I defined above, the corresponding wavefunction for real α is also a Gaussian, $\propto \exp (-(x-\sqrt{2} \alpha)^2/2) $, for t =0 --wow!