How to show the field operator creates a particle at known position in Fock space

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 $F$?
• Your inference that the LHS is in H is wrong. It is in Fock space, loosely $H^\infty$, where, given the respective field operators, you only need consider $H^2$. Just review 2nd quantization. – Cosmas Zachos May 6 '18 at 13:35
• In the course he defines $a^+$ in the context of the simple harmonic oscillator as $P+i\omega X$ where $P=-i\partial_x$ and $X: \psi\mapsto X\psi$. That seemed very much to be talking about $H$ i.e. the space spanned by $\lvert x \rangle$. The problem is that $\Psi^+(x)$ is being used for two different objects: One is a map $H\mapsto H$ and the other is a map $H^\infty \mapsto H^\infty$. He proves that for the former $\Psi^+(x)\lvert 0 \rangle = \lvert x\rangle$ but this doesn't prove it's true for the latter. – Alex Zeffertt May 6 '18 at 14:27
• It's occurred to me that one way I can proceed with the course is this: instead of attempting to construct $F$ from $H$ and $\lvert n_1n_2...\rangle \in F$ from the $\lvert x \rangle \in H$ and then prove that $\lvert x \rangle=\Psi^+(x)\lvert 0 \rangle$ is true in $F$ just like it is in $H$, I could just treat $F$ as a brand new object not derived from $H$ and take $\lvert x \rangle=\Psi^+(x)\lvert 0 \rangle$ as a definition. – Alex Zeffertt May 6 '18 at 14:35
• I think I've got it. The key fact is that $H\subset\bigoplus_nH^{\otimes n}=F$. So if $\Psi^+(x):F\mapsto F$ is defined as above, then $\Psi^+(x)|0\rangle=\sum_n\psi^*_n(x)a^+_n|0\rangle=\sum_n\psi^*_n(x)|1_n\rangle$. But $|1_n\rangle \in H^1$ and can be written $\frac{(a^+)^n}{\sqrt{n!}}|1_0\rangle$. So $\Psi^+(x)|0\rangle=\sum_n\psi^*_n(x)\frac{(a^+)^n}{\sqrt{n!}}|1_0\rangle$ which is an expression involving only operators on $H$ and vectors in $H$ and which can be shown to equal $|x\rangle$. I guess that answers the question in the title but not question no. 2 in the text. – Alex Zeffertt May 6 '18 at 15:27
• Try a decent source for introductory field theory. Using the same index, n, for a label of oscillators and for a power of a single one is bound to produce absurdities, like the one you are writing down in this last comment. I think you really mean $|1_n\rangle= a_n^\dagger|0\rangle$, but I have no idea how you connected this to the n-th excited state of a single oscillator, unsoundly... – Cosmas Zachos May 6 '18 at 16:29

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).

• In addition to what this answer says about localization, if you do try to create a perfectly localized state, you come to the conclusion that you can only localize it in one frame (i.e. localization is not a Lorentz-invariant property). – probably_someone May 7 '18 at 0:22
• Indeed. I slipped the Lorentz measure without commentary, to avoid fuss and sic the OP to the slough of despond. He appears to simply wish to appreciate the over-simplified schematic of L.S.'s presentation. – Cosmas Zachos May 7 '18 at 0:27
• In all fairness to L.S., he is basically using the Lorentz non-invariant Wigner-Newton representation without clueing his audience in. But, everyone in the audience will go on to use the above covariant representation, all their lives, and ignore the W-N with extreme prejudice... so why confuse them further? – Cosmas Zachos May 7 '18 at 13:29
• Oops. I appear to have stumbled into a war between two academics. I had a minor problem with a statement LS made in the course which is now resolved. I can see my previous comment contained errors, but the main point is that some states are in both F and H, - namely the pure single particle states - but its important not to get confused by the different labels they have when considered as a member of each space. Thank you for your answer, but I think it's going to be easier now for me to finish the course rather than change direction even if you think it's over simplified. – Alex Zeffertt May 7 '18 at 15:38
• No, no war, agreement! But, indeed, you don't care about Lorentz invariance at this stage. The most important thing to appreciate is you never go to the Hermite functions you wrote implicitly, i.e. all powers of creators. You just look at few particle states and their linear combinations. You need the infinite dimensional Fock space since your linear combinations (over k) must "sample" all oscillators... all momenta. – Cosmas Zachos May 7 '18 at 16:11