The discrepancy between first and second quantised theories (Non-relativistic) in energy and position measurements

Question

I realize I've asked a similar question before. In this question, I really want to focus on non-relativistic QM.

Energy and position measurements are straight-forward in the first quantised theory. This is because you end up with a position and an energy operator straight out of the quantisation postulate $[X, P]=i\hbar$. The position operator is part of the very definition of the theory.

But things get muddier in the second quantised non-relativistic field theory. The quantisation postulate is now applied on the field operators $[\psi (x) , \pi(y)]=i\hbar\delta^3(x-y) $. We still end up with an energy operator as the integral of the energe density $\int H(x) d^3 x$. But the position operator is not something defined in terms of local field operators. It has to be defined by laying out its eigenvectors in the momentum basis : $e^{ipx/\hbar}$ $p\in R^3$

Now this is where things get tricky. In the first quantised theory, the status of measurement devices is like omnipresent gods. But in the second quantised theory, we realize that the measurement devices occupy a small region in the same $R^3$ space that the fields occupy. So, precisely speaking, the energy that we're measuring in the second quantised theory does not correspond to the energy operator in the first quantised theory. The energy operator will be $\int _S H d^3x$ where $S$ is the local subset of the space in which the measurement device is located. In the first quantised theory, there is no operator that corresponds to this "fractional energy" operator.

This will lead to a discrepancy in the predictions of the two theories. In the first quantised theory, upon an energy measurement on the plane wave $e^{ipx/\hbar}$, we get the eigenvalue $\frac{p^2}{2m}$. But in the second quantised theory, we will only measure a fraction of that energy because the energy density of this state is spread out across the entire $R^3$ space.

So can we say that the second quantised theory is providing a more accurate description of realtity? Is it that the first quantised theory only seems to work because we only ever deal with localised systems like atoms? We never deal with highly delocalised near-plane waves in the laboratory.

The position measurement also behaves very strange in the first quantised theory. Once we detect a particle at a position $x$, the probability of detecting it at a position $y$ upon immediate successive measurements becomes $0$. This is against a tenet of second quantisation where the field operators are postulated to commute $[\psi (x), \psi (y) ]=0$. This means that a local measurement at one point shouldn't affect the expected values of the local operators at another point. Why this discrepancy between first- and second-quantised theories?

I do know that the first-quantised theory has to be abandoned after we go relatitivistic. I'm asking if the first quantised theory is strictly inaccurate even before relativity (as in, if it gives the wrong predictions for highly delocalised states)

Answer 1

We need to be much more careful with the technical details here. In the end I do not think there is any discrepancy between first- and second-quantized non-relativistic theories, but that the way we often talk about them obscures this, as the question demonstrates.

Let us first establish what the "first-quantized theory" actually is: It is not the theory of a single particle with a single position operator $x$ and a single momentum operator $p$. Instead, it is the theory of finitely but arbitrarily many such particles, i.e. it takes place on the Fock space $F(H)$ associated with the one-particle space $H$.

While it is common to present quantum field theory indeed as the quantization of a classical field theory, this is not the most useful way to think about it when trying to compare it to the first-quantized theory. Instead, we should think about it as being constructed from the first-quantized theory - the field is made out of the creation/annihilation operators on the Fock space (see e.g. Wikipedia's exposition of the Schrödinger field for a longer explanation). In the non-relativistic case, we simply have that $$ \psi^\dagger(x)\lvert x_1,\dots x_n\rangle = \lvert x,x_1,\dots,x_n\rangle$$ i.e. the adjoint of the field creates a state at $x$. The claim in the question that

But the position operator is not something defined in terms of local field operators. It has to be defined by laying out its eigenvectors in the momentum basis

is simply false - the position operator on $H$ extends straightforwardly to $F(H)$, and we can even define the field directly in position space in the non-relativistic case, with no passage through momentum space necessary.

There is a one-particle Hamiltonian operator $H_1$, and the full Hamiltonian on the Fock space is simply $$ F(H_1) = \sum_i E_i a^\dagger_i a_i,$$ where $a^\dagger_i a_i$ is the number operator for particles in the state labeled with $i$ and $E_i$ the energy eigenvalue in that state.

When these are free particles, the field Hamiltonian you get is $H = \int \frac{\nabla \psi^\dagger \nabla \psi}{2m}$, which in momentum space is $H = \int \frac{p^2}{2m} a^\dagger_p a_p$. Since the single particle Hamiltonian has eigenvalues $E_p = \frac{p^2}{2m}$, you can see that the "first-quantized" and "second-quantized" ways to get the Hamiltonian produce exactly the same energy operator (that I wrote a sum in one of them and an integral in the other is just the usually notational mess for operators with continuous spectrum).

That there is a Hamiltonian density in position space with $H = \int \mathcal{H}$ so that you can in field theory write down a "local operator" $H\rvert_S = \int_S \mathcal{H}$ in position space is in principle no different from doing something like restricting the sum in the definition of $F(H_1)$ to a subset of the indices, it's just that it's much easier to write this down in position space in the field formalism.

The second-quantized theory so far really is just a convenient repackaging of the many-body first-quantized theory, they are not different.

Lastly, you seem to be concerned about locality, but in this approach to second quantization we don't have a "tenet of field theory" that $[\psi(x),\psi(y)] = 0$ - this is just the brute fact that creation and annihilation operators for different orthogonal states commute, i.e. this is true already in the first-quantized theory, not some sort of postulate we add.

"Measuring a particle at $x$" and "Measuring a particle at $y$" in the language of the Fock space is measuring the number operators $\psi^\dagger(x)\psi(x)$ and $\psi^\dagger(y)\psi(y)$, respectively. So $[\psi(x),\psi(y)] = 0$ is true, but it doesn't directly have something to do with this measurement! Still:

\begin{align} [\psi^\dagger(y)\psi(y),\psi^\dagger(x)\psi(x)] & = \psi^\dagger(y)[\psi(y),\psi^\dagger(x)\psi(x)] + [\psi^\dagger(y),\psi^\dagger(x)\psi(x)]\psi(y) \\ & = \psi^\dagger(y)[\psi(y),\psi^\dagger(x)]\psi(x) + \psi^\dagger(x) [\psi^\dagger(y),\psi(x)]\psi(y) \\ & = \psi^\dagger(y) \delta(y-x) \psi(x) - \psi^\dagger(x) \delta(x-y) \psi(y) = 0 \end{align}

But that's not a contradiction at all because this is also true in the first-quantized theory! Measuring a particle in a region $R\subset\mathbb{R}^n$ means measuring the projection operator onto the space spanned by the position eigenstates, i.e.

$$ P_R = \int_R \lvert x\rangle\langle x\rvert \mathrm{d}x$$

in the usual Dirac notation for the spectral measure is the one-particle operator for this. When $R_1$ and $R_2$ are two disjoint regions, you have $[P_{R_1}, P_{R_2}] = 0$ (and likewise $[F(P_{R_1}),F(P_{R_2})] = 0$ as a consequence) - orthogonal projectors on orthogonal subspaces commute. So no discrepancy here, either.

That the question seems to try to play this against the measurement problem, in particular the non-unitarity or "collapse" of a measurement, suggests a confusion about the meaning of "locality" as it relates to wavefunction collapse and field theory as $[\psi(x),\psi(y)] = 0$. While we use the word "locality" or "causality" in both of these contexts, we do not mean the same concept by it. This answer is already long enough, see this question and its answers for more discussion on this point.

For a book that approaches even relativistic field theory consistently in this framework of second-quantization and the primacy of particles and their Fock space, not fields, give Weinberg's "The Quantum Theory of Fields" a read.