# Second quantization

by daudaudaudau
Tags: quantization
 P: 296 Hi. In second quantization (not QFT or anything advanced like that) we have the particle density $\hat n(x)=\Psi^{\dagger}(x)\Psi(x)$ using the usual field creation/annihilation operators. For a single particle we obtain for the expectation value in the state $|\psi\rangle$: $\langle \psi | \Psi^{\dagger}(x)\Psi(x) | \psi\rangle=|\psi(x)|^2$. So does this mean that I should think of $\Psi(x)$ as the second quantized wave function? What is the significance of this formal similarity between the wave function and the creation/annihilation operators? It seems that whenever you have something in first quantization written in terms of a wave function (i.e. the probability current density), you can replace the wave functions by the creation/annihilation operators and get the second quantized operator...
 P: 5 Yes, field operator $$\hat \Phi$$ and wave function $$\phi$$ are very similar objects. The formal reason is that $$\hat \Phi(x) = \sum_k \phi_k(x) \hat a_k$$ - it is a "superposition" of wave functions with coefficients being annihilation operators. That's why the technique is called "second quantization". In the "first quantization" we go from physical obserables (which are numbers) to operators. And now we go from wave functions (which are numbers) to operators. For example, if you write motion equations for $$\hat\Psi(x,t)$$ in the free particle theory then it will coincide with Schroedinger equation for $$\psi(x,t)$$
 P: 296 Thank you for the reply. If you have the particle density operator $\hat n(x)=\Psi^{\dagger}(x)\Psi(x)$, you get the particle density in a particular state by taking the expectation value of this operator. Is it possible somehow to get the wave function by taking the expectation value of an operator?
 Sci Advisor P: 779 Second quantization To expand slightly on SnarkyFellow's response: Think of a second-quantized field operator $$\hat{\Phi}(x)=\sum_k \phi_k(x)\hat{a}_k$$ as the "hammer that bangs the vacuum" - and everytime it bangs, it "creates" a state with wavefunction $\phi_k(x)[/tex]. Thus, when you want to talk about multiple particle states in quantum mechanics, it is useful to use this "field operator" rather than an ordinary wavefunction. That is why you use "second quantized quantum mechanics" in "many body theory". And in fact, to dau^4's first sentence: there is no real difference between this and QFT! But that is just semantics. ;-)  Sci Advisor P: 779 When [itex]\hat{n}(x)$ bangs the state you're in (analogous to you taking the expectation value of the operator), you get $|\phi_k(x)|^2$ which is, of course, the particle density, so yes!
 P: 1,160 How about the vacuum average of Ψ(x)ak+ ?
 Sci Advisor P: 779 you mean, i left out the "negative frequency" terms? sure. that's there too! i was typing quickly. I should have written: $$\hat{\Phi}(x)=\sum_k\left(\phi_k(x)a_k+\phi_k(x)^*a_k^\dag\right)$$ Then $\langle 0|\hat{\Phi}^\dag(x)\hat{\Phi}(x)|0\rangle = \sum_k|\phi_k(x)|^2$.
P: 296
 Quote by Bob_for_short How about the vacuum average of Ψ(x)ak+ ?
I guess that would do, I just think it is a little inconsistent. $\langle \psi | \Psi^{\dagger}(x)\Psi(x) | \psi\rangle$ is the particle density in the state $|\psi\rangle$. Now I want the wave function in the state $|\psi\rangle$, so I was hoping for something similarly looking.

Or maybe I'm looking for the second quantization equivalent of $\langle x|\psi\rangle=\psi(x)$, i.e in second quantization the creation or annihilation operator would be on the right and something other on the left hand side.
P: 779
 Quote by daudaudaudau I guess that would do, I just think it is a little inconsistent. $\langle \psi | \Psi^{\dagger}(x)\Psi(x) | \psi\rangle$ is the particle density in the state $|\psi\rangle$. Now I want the wave function in the state $|\psi\rangle$, so I was hoping for something similarly looking.
Not sure what you mean. The state $|\phi(x)\rangle=\hat{\Phi}(x)|0\rangle$, and the wavefunction is $\phi(x)\equiv\langle 0|\phi(x)\rangle$ in my above notation. Does that make sense?
P: 296
 Quote by blechman To expand slightly on SnarkyFellow's response: Think of a second-quantized field operator $$\hat{\Phi}(x)=\sum_k \phi_k(x)\hat{a}_k$$ as the "hammer that bangs the vacuum" - and everytime it bangs, it "creates" a state with wavefunction $\phi_k(x)[/tex]. Sorry, I don't understand your hammer-picture. The field creation operator (written as a sum, the same way you do) acting on the vacuum will create a linear combination of many different states, all weighted with the value of the corresponding wave function evaluated at x. Somehow the linear combination of all these states produce a position eigenstate. (we do agree that the field creation operator creates a position eigenstate?) P: 296  Quote by blechman Not sure what you mean. The state [itex]|\phi(x)\rangle=\hat{\Phi}(x)|0\rangle$, and the wavefunction is $\phi(x)\equiv\langle 0|\phi(x)\rangle$ in my above notation. Does that make sense?
Yeah, the only thing that bothered me was the vacuum average. I don't know if it's even possible to do what I want to do. I just wanted to see how far the formal similarities go.
 P: 296 If someone can recommend a good book about this, that would be great.. I have read Ballentine's chapters on this topic.
P: 779
 Quote by daudaudaudau Sorry, I don't understand your hammer-picture. The field creation operator (written as a sum, the same way you do) acting on the vacuum will create a linear combination of many different states, all weighted with the value of the corresponding wave function evaluated at x. Somehow the linear combination of all these states produce a position eigenstate. (we do agree that the field creation operator creates a position eigenstate?)
Consider the ordinary quantum story:

$$|\psi\rangle=\int dx\psi(x)|\psi\rangle$$

$\psi$ is not a position eigenstate, and the "wavefunction" is given by $\langle x|\psi\rangle$. My sum over k is analogous the integral over x in this past example.

So if you want to have the wavefunction of a specific mode, you would write

$$\langle k|\hat{\Phi}(x)|0\rangle=\phi^*_k(x)$$

(where the c.c. is b/c of my phase conventions). Is that better?
P: 171
 Quote by blechman Not sure what you mean. The state $|\phi(x)\rangle=\hat{\Phi}(x)|0\rangle$, and the wavefunction is $\phi(x)\equiv\langle 0|\phi(x)\rangle$ in my above notation. Does that make sense?
This makes no sense to me. Doesn't the operator $\hat{\Phi}(x)$ annihilate the particle vacuum state $$|0\rangle$$?
Moreover, the state with 0 particles is orthogonal to the one with a particle.

Perhaps there is a confusion between different second quantized notation going on. As I understand it, the notation in the original post is the following:
$\hat{\Phi}^{\dagger}(x)$ creates a particle at x. i.e, $\hat{\Phi}^{\dagger}(x)|0\rangle = | x\rangle$
$\hat{\Phi}(x)$ destroys a particle at x.

The fact that the symbol for the operators is $\Phi$ is neither here nor there, we could have chosen $c^{\dagger}_x$.

I was under the impression, that for a 1-particle system, we have the wave function when the system is in state $|\psi\rangle$ as $\psi(x) = \langle x|\psi\rangle$
Inserting the the other expression for $|x\rangle$ in second quantized language
$$\psi(x) = \langle 0|\Phi(x)|\psi\rangle$$

The quantity $$|\psi|^2$$ is then
$$|\psi(x)|^2 = \langle \psi|\Phi^\dagger(x)|0\rangle\langle 0|\Phi(x)|\psi\rangle$$
As long as $|\psi\rangle$ is a one particle state (so our single particle wavefunction makes sense), the vacuum projection operator just acts like the resolution of the identity so we can ignore it to give $$|\psi(x)|^2 = \langle \psi|\Phi^\dagger(x)\Phi(x)|\psi\rangle$$
For a many particle state, the particle density is more complicated in the wavefunction picture, but the second quantised operator remains the same.

As far as book recommendations go, if you are coming at this from the condensed matter perspective, the book by Negele and Orland is highly recommended (by me at least) - the first couple of chapters on second quantisation are very thorough. "Condensed matter field theory" by Altland and Simons has a good intro to second quantisation - you can get the flavour of the book by reading Ben Simon's lecture notes for his Quantum Condensed Matter Field Theory course. (google him). MIT Open Course Ware might well have some decent lecture notes available too, I haven't checked. Wen has a recent graduate level Condensed Matter Field Theory textbook too, which surely has a chapter on second quantization.
 Sci Advisor P: 779 i think, as you say, this is notational. but if you look at my earlier post, i have the field operator as the SUM of creation and annihilation operators, so this works itself out. If you split it up as positive and negative frequency terms, then you are right. I also used Negele and Orland, an excellent book, but it is quite advanced (at least I think so).
P: 296
Thank you for clearing things up, Peter, and I will definitely have a look at those books. Could you maybe explain a little more about why the vacuum projector becomes the identity operator in the single particle case?

 Quote by peteratcam The quantity $$|\psi|^2$$ is then $$|\psi(x)|^2 = \langle \psi|\Phi^\dagger(x)|0\rangle\langle 0|\Phi(x)|\psi\rangle$$ As long as $|\psi\rangle$ is a one particle state (so our single particle wavefunction makes sense), the vacuum projection operator just acts like the resolution of the identity so we can ignore it to give $$|\psi(x)|^2 = \langle \psi|\Phi^\dagger(x)\Phi(x)|\psi\rangle$$
 P: 171 A full identity operator (for Fock space) would have been $$|0\rangle\langle 0| + \sum_k|k\rangle\langle k| + \sum_{k_1,k_2}|k_1,k_2\rangle\langle k_1,k_2| + \ldots$$ I.e, the 0-particle identity $$\oplus$$ the identity for a one particle state $$\oplus$$ the identity operator for two particle states etc. I've chosen 'k' to be the labels of a basis of single particle states. The creation (annihilation) operators move us from an n-particle state to an n+1 (n-1) particle state. States with different numbers of particles are orthogonal. So consider: $$\langle\psi|\Phi^{\dagger}(x)|S\rangle$$ If $$\langle \psi|$$ is a one particle state, then the only state S for which the above is non-zero is a zero particle state, and there is only one of those, the particle vacuum. So inserting a full resolution of the indentity would reduce to exactly the expression I gave.
 P: 5 As peteratcam has noticed, wavefunction function can be obtained as matrix element $$\langle 0|\hat \Psi |\psi_k \rangle$$. We can't get a wavefunction as the expectation in a state with defined number of particles. It's due to the fact that $$\hat \Psi$$ decreases number of particles so in order to give nonzero result, there should be the product of equal number of creation and annihilation operators. But such a product is invariant under the gauge transformation $$\phi \rightarrow \phi e^{i\alpha}$$. Therefore it can't give us the wavefunction. But if we prepare somehow $$\alpha|0\rangle+\beta|\phi\rangle$$ state, then expectation of $$\hat \Psi$$ is $$\alpha^*\beta \phi$$. It's rather curious fact, but I suppose it is not really useful in physics because $$\hat \Psi$$ isn't self-conjugate on Fock space. Therefore it is not observable and we can't get this expectation in an experiment.

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