# Second quantization

1. Dec 3, 2009

### daudaudaudau

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...

2. Dec 4, 2009

### Snarky Fellow

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 $$\psi(x,t)$$

3. Dec 4, 2009

### daudaudaudau

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?

4. Dec 4, 2009

### blechman

To expand slightly on SnarkyFellow's response:

Think of a second-quantized field operator

$$\hat{\Phi}(x)=\sum_k \phi_k(x)\hat{a}_k$$

6. Dec 4, 2009

### Bob_for_short

How about the vacuum average of Ψ(x)ak+ ?

7. Dec 4, 2009

### blechman

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 $\langle 0|\hat{\Phi}^\dag(x)\hat{\Phi}(x)|0\rangle = \sum_k|\phi_k(x)|^2$.

8. Dec 4, 2009

### 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.

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.

Last edited: Dec 4, 2009
9. Dec 4, 2009

### 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?

10. Dec 4, 2009

### 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?)

11. Dec 4, 2009

### daudaudaudau

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.

12. Dec 4, 2009

### daudaudaudau

13. Dec 4, 2009

### blechman

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?

14. Dec 4, 2009

### peteratcam

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.

15. Dec 4, 2009

### blechman

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).

16. Dec 4, 2009

### daudaudaudau

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?

17. Dec 4, 2009

### peteratcam

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.

18. Dec 5, 2009

### Snarky Fellow

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.

Last edited: Dec 5, 2009
19. Dec 5, 2009

### peteratcam

This started me thinking, so in case it is of benefit to anyone else, I'll think out loud. I just wanted to check that the wavefunction/second quantisation correspondence works for many particle states equally well:

Write the Fock space identity using a position basis:
$$|0\rangle\langle 0| + \int dx|x\rangle\langle x| + \int dx_1 dx_2|x_1,x_2\rangle\langle x_1,x_2| + \ldots$$

Now consider the particle density in a two particle state $|\psi\rangle$:
$$\rho(x) = \langle \psi|\Phi^\dagger(x)\Phi(x)|\psi\rangle$$
Insert two copies of the resolution of the identity. By orthogonality of state of different particle number, we get:
$$\rho(x) = \int dx_1 dx_2dx_1' dx_2'\langle \psi|x_1,x_2\rangle\langle x_1,x_2|\Phi^\dagger(x)\Phi(x)|x_1',x_2'\rangle\langle x_1',x_2'|\psi\rangle$$

Now, the matrix element in the middle $$\langle x_1,x_2|\Phi^\dagger(x)\Phi(x)|x_1',x_2'\rangle$$ is trivial because we are in the diagonal basis for $$\Phi^\dagger(x)\Phi(x)$$.

$$\Phi^\dagger(x)\Phi(x)|x_1',x_2'\rangle = [\delta(x-x_1') + \delta(x-x_2')]|x_1',x_2'\rangle$$
So the matrix element evaluates as:
$$[\delta(x-x_1') + \delta(x-x_2')]\langle x_1,x_2|x_1',x_2'\rangle=[\delta(x-x_1') + \delta(x-x_2')]\delta(x_1-x_1')\delta(x_2-x_2')$$

Inserting this into the expression for the particle density, recognising the many-particle wavefunction, and integrating over the primed coordinates gives:
$$\rho(x) = \int dx_1 dx_2|\psi(x_1,x_2)|^2[\delta(x-x_1) + \delta(x-x_2)]$$
which is exactly the particle density in good old wavefunction language, since the particle density operator in normal quantum mechanics is:
$$\hat n(x)=\sum_i\delta(x-\hat x_i)$$
Why don't we normally see this operator much in normal QM?
It's because we normally have the situation that we want to assign a potential V(x) when the particle is found at x, for all x:
$$\int V(x)\hat n(x) dx = \sum_i V(\hat x_i)$$
the RHS is of course the common way of writing a potential term in a many body first quantised hamiltonian.

20. Dec 5, 2009

### daudaudaudau

I wonder how to derive this. For a single particle I have $\Phi^{\dagger}(x)\Phi(x)|x'\rangle$. I integrate this over all $x$ to get the number operator
$$\int dx\Phi^{\dagger}(x)\Phi(x)|x'\rangle=\hat N|x'\rangle=|x'\rangle$$

So I guess this shows that $\Phi^{\dagger}(x)\Phi(x)|x'\rangle=\delta(x-K)|x'\rangle$. The expression has to be zero for all $x\neq x'$, so we have $K=x'$. My only issue with this is that I would expect that $\Phi^{\dagger}(x')\Phi(x')|x'\rangle=|x'\rangle$, i.e. you remove the particle and add it again yielding the initial state and no delta function.

To me the density operator $\hat n(x)$ looks like it tells us that we have particles in the specific positions $x_i$, but this seems wrong because in general we don't know the exact positions of the particles. Also, I'm thinking of the particle density as analogous to the probability density, but you associate it with a potential?

Last edited: Dec 5, 2009