quantum mechanics - Basic question on holomorphic formalism in QM

In my course, the teacher introduced us the holomorphic formalism in Quantum Mechanics.

What I basically understood is that initially, we work in the Hilbert space of square integrable functions $\mathcal{H}_n$.

I will write a function from this space $\psi(q)$.

When we work with harmonic oscillator, we construct the following operator :

$$a=\frac{q+ip}{\sqrt{2}}$$

It helps us to diagonalise the Hamiltonian.

So, from what I have understood of the holomorphic formalism is that we want to associate to each wavefunction from $\mathcal{H}_n$ an holomorphic function living in the space $\mathcal{F}_n$.

To do it we need different things:

• Create a scalar product on $\mathcal{F}_n$


• Find the function $A: \psi(q) \mapsto f(z)$ that will tell me how I translate my wavefunction from $\mathcal{H}_n$ into the holomorphic space $\mathcal{F}_n$.

We write in the holomophic space:

$$\langle f,g \rangle = \int \bar{f}(z)g(z)\rho(x,y)dxdy$$

Thus, we want to find the function $\mathbb{\rho}$.

What I understand from the course is that:

As we have $[a_k, a_l^{\dagger}]=\delta_{kl}$ in $\mathcal{H}_n$, and $[\frac{\partial}{z_k}, z_l]=\delta_{kl}$ in $\mathcal{F}_n$, the "corresponding" operator to $a_k$ will be $\frac{\partial}{z_k}$.

And we need to build the scalar product such that $z_l$ is the hermitic conjugate of $\frac{\partial}{z_k}$.

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My questions (probably very basic and obvious...)

Why is it enough to find operators in the holomorphic space that has the same commutation relations than the one in the wavefunctions space ? Is it because it is enough to ensure a "bijection" between the two spaces ?

What I mean is that if I do the following operations:

$$\begin{array}{ccc} & & \\ \hline \psi(q) & \rightarrow & F(\widehat{a},\widehat{a}^{\dagger})\psi(q) &\\ \downarrow & & \updownarrow \\ f(z) & \rightarrow & F(\frac{\partial}{\partial z},z)f(z)=g(z) & \end{array}$$

If I am in the wavefunction space, I apply a function F of the operators $a, a^\dagger$, and then I go to the holomorphic space, I will find a function $g(z)$.

Now if I initially go into the holomorphic space and apply the same function F but depending on the corresponding operators $\frac{\partial}{\partial z}$, $z$, I will end up with the same function $g(z)$.

So in a sense because the operators follows the good commutations relations I will always land on my feet at the end.

Another question:

We could imagine any other operator acting on the holomorphic space following the same commutation relation to work with? The only thing that would change with them would be our scalar product $\rho(x,y)$ right?