Completeness relation for coherent states of the quantum harmonic oscillator

Question

For the Quantum harmonic oscillator with energy eigenstates $|n\rangle$ one defines a coherent state for every complex number $z$ by setting (note that the normalization varies across the literature) $$\psi_z:=\sum_{n=0}^{\infty} \frac{z^n}{\sqrt{n!}}|n\rangle.$$ In the book "Quantum Theory - a mathematical approach" by Peter Bongaarts he claims the following result without proof (I slightly changed the notation):

Any vector $\phi$ in the Hilbert space (of the oscillator) can be expressed as $$\phi=\frac{1}{\pi}\int_{z\in\mathbb{C}}F_{\phi}(z)e^{-|z|^2}\psi_z\qquad \tag{1}$$ where the function $F_\phi$ is uniquely determined by $$F_\phi(z)=\langle \psi_z, \phi \rangle.$$ This function is anti-holomorphic and satisfies $$\int_{z\in\mathbb{C}}|F_{\phi}(z)|^2 e^{-|z|^2}<\infty.$$

Basically what I'm trying to do is to prove this claim. It is rather easy to see (ignoring questions like "what do I need to know about Bochner Integrals to change order with infinite sums"), that taking the specified function $F_\phi$ one does get equality (1) and also to show that it is anti-holomorphic.

However, I do not believe that the function is necessarily unique among ALL measurable functions $\mathbb{C}\to\mathbb{C}$. Best case scenario would be that two such functions differ on a set of measure zero, but even that seems unlikely. I suspect the function is unique among the anti-holomorphic functions. Does anyone know the correct claim for this and the proof or where to find it? The next thing I'm struggling with is to show that the last integral is bounded.

I'm very grateful for any advice.

Answer 1

I wanted this post to be a comment, but made it an answer instead. I cant delete it yet, so i'll extend it a bit.

You might want to check Quantum mechanics for mathematicians by L. Takhtajan; the result is also in this book, but I don't remember if he proves it.

In any case, you are right in that the uniqueness only holds for antiholomorphic functions. Roughly speaking the argument is as follows: the finite dimensional spaces of the $n$ lowest harmonic oscillator eigenstates are list of $n$ complex numbers. These are in $1$-$1$ correspondence with antiholomorphic polynomials. By taking the metric closure on both sides, one obtains the required isomorphism between vector spaces.

You can check that the isomorphism you give above sends $|n\rangle$ to the monomial $\bar z$, up to a constant, in agreement with what I just said.

Answer 2

I will try to answer the integral question.

A continuous function on a bounded interval is bounded.

Throughout this answer I will use Riemann's definition for integrability.

Using Cauchy-Schwarz Inequality-

\begin{equation} \int_{z\in\mathbb{C}}|F_{\phi}(z)|^2 e^{-|z|^2}\leq\Bigg|\int_{z\in\mathbb{C}}|F_{\phi}(z)|^2 e^{-|z|^2}\Bigg|\leq\sqrt{\int_{z\in\mathbb{C}}|F_{\phi}(z)|^4}\sqrt{\int_{z\in\mathbb{C}} e^{-2|z|^2}} \end{equation}

\begin{equation} |F_{\phi}(z)|=|\langle \psi_z, \phi \rangle|\leq|\langle \psi_z\rangle||\langle\phi \rangle| \end{equation}

The wave functions in quantum mechanics must always be continuous differentiable and must always be square integrable.

As a result $|\langle \psi_z\rangle||\langle\phi \rangle|$on a bounded interval is also bounded.

Then $|F_{\phi}(z)|^4$ is also bounded and consequently $$\sqrt{\int_{z\in\mathbb{C}}|F_{\phi}(z)|^4} <\infty$$.

The function $f(z)=e^{-2|z|^2}$ is real (assuming $|z|$ is the modulus). And thus is continuous and consequently bounded for any $x, y\in\mathbb{R}$, where $z=x+iy$.

As a result the second integral is also bounded for any $z\in\mathbb{C}$.

This means that $$\sqrt{\int_{z\in\mathbb{C}}|F_{\phi}(z)|^4}\sqrt{\int_{z\in\mathbb{C}} e^{-2|z|^2}}$$ is bounded.

And using the inequality above we get that the original integral is also bounded.