Quantum fields as differential operators

Question

As I understand it, there was initially two formalism for QM, before Dirac reunites them both with his famous braket notation:

• Schrödinger's formalism that involved differential operators acting on wave functions,

• Heisenberg's formalism that involved linear operators acting on vectors.

Now, if we consider a scalar field $\phi$, the quantum field $\hat{\phi}$ is an operator so it acts on kets. We have an explicit expression of $\hat{\phi}$ in terms of the annihilation and creation operators $a_{\mathbf {p} }$ and $a_{\mathbf {p} }^\dagger$:

$$ \hat{\phi} (\mathbf {x} ,t)=\int {\frac {d^{3}p }{(2\pi )^{3}}}{\frac {1}{\sqrt {2\omega _{\mathbf {p} }}}}\left(a_{\mathbf {p} }e^{-i\omega _{\mathbf {p} }t+i\mathbf {p} \cdot \mathbf {x} }+a_{\mathbf {p} }^\dagger e^{i\omega _{\mathbf {p} }t-i\mathbf {p} \cdot \mathbf {x} }\right).$$

The creation and annihilation operators come from creation and annihilation operators of the harmonic oscillator. These can be expressed in terms of position and momentum operators, which have an expression and term of differential operators. So is there a way to view $\hat{\phi}$ like a differential operator acting on wave functions (like Schrödinger's formalism)? In the same way as $\hat{\mathbf{P}} = -i\hbar\nabla$ for example? Is there any literature about this?

Answer 1

Well, given a field-theoretic equal-time CCR $$ [\hat{\phi}({\bf x},t),\hat{\pi}({\bf y},t)]~=~i\hbar {\bf 1}~ \delta^3({\bf x}\!-\!{\bf y}),$$ there exists the corresponding Schrödinger representation $$\hat{\pi}({\bf x},t)~=~\frac{\hbar}{i}\frac{\delta}{\delta \phi({\bf x},t)}, \qquad \hat{\phi}({\bf x},t)~=~\phi({\bf x},t),$$ which writes the momentum field $\hat{\pi}({\bf x},t)$ as a functional derivative. See also the Schrödinger functional.

Answer 2

Not every operator can be represented as a differential form - spin is a good example.

However, the difference between Heisenberg and Schrödinger formalism was not that of differential operators versus matrices. Schrödinger built a consistent quantum mechanical picture based on wave equation (that received his name) - wave mechanics, whereas Heisenberg built matrix mechanics, where the dynamics was described by the Heisenberg equation of motion for operators. The difference is akin to that between the Hamilton-Jacobi equations and Poisson brackets in classical mechanics. Quantum mechanics still rather faithfully refers to this distinction by using terms Schrödinger picture and Heisenberg picture for the situations where the time dependence is carried by wave functions and operators respectively.

Answer 3

I think I know what you are asking so I will answer with some rough ideas that might help you get some insight.

Every separable Hilbert space is isometrically isomorphic to a space $L^{2}(\mathbb{R}^n)$. If I have a separable Hilbert space $X$, let $i:X\rightarrow L^2(\mathbb{R^n})$ be the isomorphism. If $A$ is an operator on $X$ then $A'$ is an operator on $L^2 (\mathbb{R}^n)$ where $A' = i A i^{-1}$. This gives a correspondence between operators on abstract vectors and operators on functions.

The main difference between the Heisenberg and Schrodinger pictures is that in Heisenberg, we view the operators as changing over time while in the Schrodinger picture the operators are fixed and the states themselves are time-dependent. In other words, in the Heisenberg picture we have some fixed state space $X$ and we have some operators $A(t)$ that act on it. The $A(t)$ is a group representation in the sense that $A(t + s) = U(s)A(t)U(s)^{-1}$ for some unitary transformation $U(s)$ which depends smoothly on $s$. In quantum field theory we move from one dimension of time to four dimensions of space-time. Hence we should have operators transform as $A(x^{\mu} + s^{\mu}) = U(s^{\mu})A(x^{\mu})U(s^{\mu})^{-1}$.

In the Schrodinger picture we think of states as having time dependence. So, there is some curve $\psi: \mathbb{R} \rightarrow X$ representing the evolution of the state over time. This also has to transform unitarily so that $\psi(t + s) = U(s) \psi(t)$. Now the operators are seen as fixed. In the QFT scenario we can still treat operators as fixed, and think of the states as having both space and time dependence. I will write $\psi(x^{\mu})$ but don't confuse this with the conventional wave function. For a given point in space-time, $\psi(x^{\mu})$ is itself an abstract vector in a separable Hilbert space. Or we may equivalently view it as a function on some geometric space (such as the space of displacements of a harmonic oscillator). We have some unitary representation U of the Lorentz group so that $\psi(x^{\mu} + s^{\mu}) = U(s^{\mu})\psi(x^{\mu})$. Now the momentum and position operators are fixed operators.

Traditional QFT actually uses something like the Heisenberg picture I stated above. However it is possible to formulate it such that a "field" is seen as a function taking space time points as inputs and functions as outputs. This is perhaps more like a classical field. It assigns a wavefunction to each point in spacetime. Then the momentum and position operators act pointwise at each location.