3
$\begingroup$

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 $\frac{\partial}{z_k}$.


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?

$\endgroup$
2
$\begingroup$

The reason that you can pass from the holomorphic representation to the real representation and vice versa is the existence of a unitary transformation between the two representations. This transformation is called the Segal–Bargmann transform given in your notation by (From the Wikipedia article): $$ f(z) = (B\psi)(z) = \int e^{-z^2+2 \sqrt{2} z q - \frac{q^2}{2}} \psi(q) dq$$ Its inverse is given by: $$ \psi(q) = (B^{-1}f)(q) = \int e^{-\bar{z}^2+2 \sqrt{2}\bar{z} q - \frac{q^2}{2}} f(z) dz d\bar{z}$$ When this transformation acts on the Hermite eigenfunctions of the Harmonic oscillater it produces the monomials $z^n$, for the $n$-th energy level in the holomorphic representation.

In geometric quantization, these choices of representations are called choices of polarizations. In the real polarization, the wave function is required to depend only on the position and not the momentum: $$\frac{\partial}{\partial p} \psi(p, q) = 0$$ In the holomorphic quantization $$\frac{\partial}{\partial \bar{z}} f(z, \bar{z} ) = 0$$ The need of a polarization stems from the fact that the space of functions of the entire phase space is too large to obtain irreducible representations of the operator algebra. In quantum mechanics of elementary systems, we require irreducibility according to Dirac's axioms.

In the general case, when the phase space is not flat, the existence of these types of polarization is not automatic. Sometimes they both exist, sometimes only one of them exists, and sometimes they do not exist. When they both exist, their equivalence is also not automatic. Also, the proofs of the equivalence of the two types of polarization are quite involved. Please see the following work by Nunes where he summarizes some of his results on the equivalence of polarizations for certain types of phase spaces.

$\endgroup$
1
$\begingroup$

Your question is not specific to the holomorphic representation at all - you could equally well ask why it suffices to observe that multiplication and differentiation obey the commutation relations of position and momentum to determine that they "are" the position and momentum operators.

The actual answer compels us to think a bit more carefully about what we are actually doing when we do quantum mechanics. Although there are certainly different viewpoint, in this case it is the algebraic viewpoint that clears up the confusion of "different Hilbert spaces" and how we "identify" correct operators. I also discuss this in this answer of mine about whether or not "the state space is somehow defined by the observables". For slightly more details on what follows please see there.

The short story is that we "start" quantum mechanics not with any Hilbert space given at all, but merely with an abstract $C^\ast$-algebra of observables. This is where the "abstract" commutation relations like $[x,p] = \mathrm{i}\hbar$ or $[a,a^\dagger] = 1$ live. $x,p,a,a^\dagger$ are not operators on any Hilbert space here, just elements of our abstract $C^\ast$-algebra. A quantum-mechanical state is a certain kind of linear functional on this algebra, and it turns out (cf. GNS construction) that for every such state there exists a Hilbert space and a representation map from the abstract algebra to the algebra of operators on that Hilbert space such that the state is associated to a vector in that Hilbert space, and that the functional is simply given by taking expectation values with respect to that state in the standard fashion. This construction obviously also works in the reverse direction: Given a Hilbert space with a representation of our algebra of observables, the vectors in it define states in precisely that fashion.

So if we find some operators on a Hilbert space that fulfill precisely the same commutation relations as our abstract observables, that means we can just map the abstract operators to these concrete operators and have a representation of the $C^\ast$-algebra, with all the vectors in that Hilbert space being proper quantum-mechanical states.

So yes, in principle you can choose whatever Hilbert space you like, with whatever inner product you like, as long as you know which operators form the representation of your algebra of observables by having the correct commutation relations. There's a caveat here that while all infinite-dimensional separable Hilbert spaces are isomorphic as Hilbert spaces, not all representations of all algebras of observables on such spaces are isomorphic. Whether they are depends on the specific algebra: The standard algebra of $[x_i,p_j] = \mathrm{i}\hbar\delta_{ij}$ (and therefore also of the associated $a_i,a_i^\dagger$) is guaranteed to have just a single unique representation by the Stone-von Neumann theorem, but if there are infinitely many $x_i$ (as in quantum field theory, essentially), then there are uncountably many non-isomorphic representations, an issue at the heart of Haag's theorem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.