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I'm very confused by something I saw in Susskind's Advanced Quantum Mechanics Lecture 6. He introduces Fock space $F$, defines the creation/annihilation operators $a^+_n,a^-_n$ on it (in terms of their action on the basis states $\lvert n_1n_2...\rangle$) and then defines the field operator on $F$ $$ \Psi^+(x) = \sum_n \psi_n^*(x) a^+_n $$ where $\psi_n(x)$ is the nth energy eigenfunction for the single particle space $H$. He then attempts to show that $\Psi^+(x)\lvert 0 \rangle$ is the state with one particle at position $x$, that $\Psi^+(y)\Psi^+(x)\lvert 0 \rangle$ is the state with two particles at positions $x$ and $y$, and so on. Unfortunately I can't follow the reasoning.

First he shows that if $a^+,a^-$ are the ladder operators on the single particle space $H$, and $\Psi^+(x) = \sum_n \psi_n^*(x) (a^+)^n$ then $\Psi^+(x)\lvert 0 \rangle = \lvert x\rangle$, i.e. this operator on $H$ maps the lowest energy eigenstate to the state of know position $x$. I followed that bit. But then he goes on to say that since $\Psi^+(x), \Psi^+(y)$ commute that

$$ \Psi^+(y)\Psi^+(x)\lvert 0 \rangle = \lvert x,y\rangle $$

But this makes no sense (to me) because the LHS is a state in $H$ and the RHS is a state in $H^2$. So my questions are:

  1. Does this last equation actually mean something?
  2. How do we achieve the real goal, which is to show that this equation is true in the fock space $F$?
Cosmas Zachos
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Alex Zeffertt
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1 Answers1

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You may well read up in Ch 2 of Tong's notes. I really shouldn't do forensic reconstruction of Lenny's lecture avoiding his pitfalls. Below, I'll give you some QM vs QFT standard cautionary contrast points... things not to do, paths not to take... you succeeded in taking most of them. Since half the misconceptions are due to notation that invites you to project analogies that simply aren't there, I'll use aggressively different symbols, and it is up to you to superpose your notation on them.

In QM, for one oscillator, x is the eigenvalue of the operator $\hat x$, and corresponds to the excursion of the system from the equilibrium point. Indeed, there is an operator that maps the ground state to the position eigenstate, both normalized, $$ \bbox[yellow]{\frac{e^{x^2/2}}{\pi^{1/4}} e^{-\bigl (a^\dagger-\sqrt{2} x\bigr )^2/2}|0\rangle=|x\rangle }, $$ so that , $$ \frac{(a+a^\dagger)}{\sqrt 2} ~ \frac{e^{x^2/2}}{\pi^{1/4}} e^{-(a^\dagger-\sqrt{2} x)^2/2} |0\rangle= x~ \frac{e^{x^2/2}}{\pi^{1/4}} e^{-(a^\dagger-\sqrt{2} x)^2/2} |0\rangle ~. $$

If you fussed, you could see that this would reduce to a superposition of Hermite functions, $$ \psi_n(x)= \langle x|n\rangle , $$ the n-th level energy eigenfunctions of the oscillator hamiltonian, acting on the corresponding excited states, so $$ |x\rangle= \sum_n \psi^*_n(x) |n\rangle ~, $$ as Lenny writes.

So, x has turned to a parameter, and one may ignore what it started out meaning.

In QFT, we package together an infinity of oscillators. I would advise against thinking of them as living far away from each other, and stick to the conventional conceit that x is a label parameter vaguely corresponding to the location of a chain of coupled oscillators, which have been diagonalized to normal modes, labelled by k. You may think of these ks as a phonon momentum label, and x as a Fourier conjugate variable of this label. But x has nothing to do with excursions from equilibrium positions or any actual operator/dynamical connection to the commutation relation of the $a_k, a_k^\dagger$. (If anything, it is the classical analog of the field below that corresponds to excursions from equilibrium.) You forget this, and you have brought tears on yourself.

A quantum field is a linear combination of these decoupled normal mode oscillators, so, for a free quantum field, it is a hermitean operator $$ \Phi (x)\sim \int dk \frac{1}{\sqrt[4]{k^2+m^2}} (e^{ikx} a_k +e^{-ikx} a_k^\dagger ). $$ These fields and their conjugate momenta (cf. Tong, or whomever) obey interesting commutation relations with interesting normalizations, involving the label k, a parameter m, etc, ... ignore for the present purposes.

Indeed, acting on the vacuum with this operator, you got yourself something formally reminiscent of the above single oscillator expression, $$ \Phi (x)|0\rangle \sim \int dk \frac{e^{-ikx}}{\sqrt[4]{k^2+m^2}} a_k^\dagger |0\rangle = \int dk \frac{e^{-ikx}}{\sqrt[4]{k^2+m^2}} |1_k\rangle , $$ in infinite-dimensional Fock space, now, where I have skipped the ground state of every other oscillator but the k excited by just one creator. Meh! you may choose to think of the x-dependent coefficient as some sort of free wavefunction of a plane wave, and, indeed, you'd get sines and cosines in a box, etc... But nowhere does Lenny insinuate these have anything to do with Hermite functions, as you were concluding!

As for the localization, no, that's another slough of despond! It sort of localizes around x, but it is not an exact δ-function as some assume at a price. $\langle 0|\Phi(x) \Phi(y) |0\rangle$ is a conventional Fock space object to compute, and is a modified Bessel function, highly peaked about coincident x and y.

Now to your questions 1. and 2. Lenny keeps the creation half of Φ(x) which creates a first excitation ("particle) centered on x, and then likewise for Φ(y), creating another excitation centered on y. Formally, it is something like $$ \Phi(y)^+\Phi (x)^+|0\rangle \sim \int dk dk' \frac{e^{-ikx-ik'y}}{\sqrt[4]{(k^2+m^2 )(k' ^2+m^2)}} |1_k, 1_k'\rangle , $$ where recall the labels k,k' correspond to plane-wave momenta, and all other labels have 0 occupation number in Fock space. Naturally, when k and k' coincide, we have a double excitation, $|2_k\rangle$, instead. So, these states amount to an infinity of oscillators in the ground state, and two in the first excited state or one in the second excited state, linearly combined with their peers of different k labels (momenta).

Cosmas Zachos
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