# Second quantization

## Main Question or Discussion Point

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...

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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 $$\psi(x,t)$$

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?

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 $\phi_k(x)[/tex]. Thus, when you want to talk about multiple particle states in quantum mechanics, it is useful to use this "field operator" rather than an ordinary wavefunction. That is why you use "second quantized quantum mechanics" in "many body theory". And in fact, to dau^4's first sentence: there is no real difference between this and QFT! But that is just semantics. ;-) blechman Science Advisor When [itex]\hat{n}(x)$ 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!

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

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

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.

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blechman
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?

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 $\phi_k(x)[/tex]. Sorry, I don't understand your hammer-picture. The field creation operator (written as a sum, the same way you do) acting on the vacuum will create a linear combination of many different states, all weighted with the value of the corresponding wave function evaluated at x. Somehow the linear combination of all these states produce a position eigenstate. (we do agree that the field creation operator creates a position eigenstate?) Not sure what you mean. The state [itex]|\phi(x)\rangle=\hat{\Phi}(x)|0\rangle$, 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.

blechman
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?

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.

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).

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?

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 $$|\psi(x)|^2 = \langle \psi|\Phi^\dagger(x)\Phi(x)|\psi\rangle$$

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.

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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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.

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 $$\Phi^\dagger(x)\Phi(x)$$.

$$\Phi^\dagger(x)\Phi(x)|x_1',x_2'\rangle = [\delta(x-x_1') + \delta(x-x_2')]|x_1',x_2'\rangle$$
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.

$$\hat n(x)=\sum_i\delta(x-\hat x_i)$$
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.
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?

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The thing with the delta functions is because of the way position states form a continuous basis, and delta functions start appearing everywhere.

The conventional normalisation in this case is $$\langle x' | x \rangle = \delta(x-x')$$. (Notice that the d-dimensional delta function has dimensions 1/L^d, so continuous states are not normalised in the same way as discrete states)

The number operator which is associated with a continuous basis is really a number density operator.

So when you expect
$\Phi^{\dagger}(x')\Phi(x')|x'\rangle=|x'\rangle$
if you think of the operator as a number density, then it seems more natural to have
$\Phi^{\dagger}(x')\Phi(x')|x\rangle=\delta(x-x')|x\rangle$

Another way to think is that the states $$|x\rangle$$ have the 'wrong' dimension, having a dimension of 1/sqrt(L^d) for a d dimensional space.
A sensible state is made up of $$|x\rangle$$, only under an integral.
Typically:
$$|\psi\rangle = \int d^d x \psi(x) |x\rangle$$
To interpret $$|\psi|^2$$ as a probability density function, it must have dimensions 1/L^d. Since the state vector $$|\psi\rangle$$ should be considered dimensionless, we must have $$|x\rangle$$ with dimensions 1/sqrt(L^d).

With more particles, you get more delta functions hanging around.

With regards to your second question, it is important to distinguish between the particle density, and the particle density operator. Just as it is important to distinguish between the position, and the position operator. I've tried to do that with hats on the operators.

The particle density and probability density are basically the same yes. In writing down a hamiltonian for a system you need the particle density operator to deal with interactions with external potentials, which is why I mentioned it.

Ultimately, there must be some operator which can be written in first quantised form which gives the particle density at postion x, and I've given you what it is. In the definition, the delta function of an operator obviously looks a bit dodgy, but we're physicists and we can get away with that.

The number operator which is associated with a continuous basis is really a number density operator.

So when you expect
$\Phi^{\dagger}(x')\Phi(x')|x'\rangle=|x'\rangle$
if you think of the operator as a number density, then it seems more natural to have
$\Phi^{\dagger}(x')\Phi(x')|x\rangle=\delta(x-x')|x\rangle$
Yes, that makes sense. I went wrong when I assumed that $\Phi(x)|x\rangle=|0\rangle$. For a general state you must have $\Phi(x)|\psi\rangle=\psi(x)|0\rangle$ and in the limit $|\psi\rangle\rightarrow|x'\rangle$ you get the delta function in place of $\psi(x)$.

Another way to think is that the states $$|x\rangle$$ have the 'wrong' dimension, having a dimension of 1/sqrt(L^d) for a d dimensional space.
A sensible state is made up of $$|x\rangle$$, only under an integral.
Typically:
$$|\psi\rangle = \int d^d x \psi(x) |x\rangle$$
To interpret $$|\psi|^2$$ as a probability density function, it must have dimensions 1/L^d. Since the state vector $$|\psi\rangle$$ should be considered dimensionless, we must have $$|x\rangle$$ with dimensions 1/sqrt(L^d).
That makes sense because the inner product between two discrete states is a probability amplitude, and between two continuous states is a probability density amplitude. Funny though that some states have a dimension while others don't...

With regards to your second question, it is important to distinguish between the particle density, and the particle density operator. Just as it is important to distinguish between the position, and the position operator. I've tried to do that with hats on the operators.
Yes, you are using operators in places where I would usually just put numbers, e.g. $x$ instead of $\hat x$. Of course you are doing it right and I am assuming to be in the position representation :)

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.
By the way, what does it mean to use this "direct sum" $\oplus$ here? I can see that if you want to sum the different identity operators, you can't use a regular sum because they have different dimensions. If we were not dealing with Fock space but with a two particle state such as $|\psi\rangle=|\phi\rangle\otimes|\varphi\rangle$, the identity operator would be given by the "direct product" of the respective identity operators, right?

The wikipedia entry on Fock space has it correct: Fock space is a direct sum of tensor products of single particle Hilbert spaces. What you say sounds right too, the identity for multiparticle states is a direct product of single particle identities. Although I suppose one has to be slightly careful about symmetrisation for multiparticle states of the same particle, so really there is a (anti)symmetrised direct product of single particle identities - not sure about this, the beauty of second quantisation is that the notation does the hard work for you.