What is the normalization of photon states in QFT?

[TriTertButoxy]

If you inspect the QED lagrangian, you'll find that the electromagnetic vector potential must have the units of [1] energy.

According to Peskin and Schroder's QFT text, the expansion of the electromagnetic field operator is

$$A_\mu(x)=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf{p}}}}\sum_{r=0}^3\bigg(a_{\mathbf{p}}^r\epsilon_\mu^r(p)e^{-ip\cdot x}+a_{\mathbf{p}}^{r\dagger}\epsilon_\mu^{r*}(p)e^{ip\cdot x}\bigg).$$​

3 factors of energy from $d^3p$, and -1/2 factor from $1/\sqrt{2E_{\mathbf{p}}}$. If the polarization vectors, $\epsilon_\mu^r$, are unitless, then the ladder operators must carry units of 1½.

How are the photon states normalized?

$$\langle0|a_{\mathbf{p}}^ra_{\mathbf{p}}^{r\dagger}|0\rangle=?$$​

 

[Timbuqtu]

I suppose you meant -1½, because 1 = 3 - ½ - 1½. You could 've seen this by looking at the commutation relation: $$\left[a_{\mathbf{p}}^r,a_{\mathbf{p'}}^{r\dagger}\right]=\delta^3(\mathbf{p}-\mathbf{p'})$$. Notice that the Dirac delta has units -3.

Plugging this relation into the vacuum we get: $$\langle0|a_{\mathbf{p}}^ra_{\mathbf{p'}}^{r\dagger} |0\rangle=\delta^3(\mathbf{p}-\mathbf{p'})$$

 

[TriTertButoxy]

Oh! I feel so stupid...

...ok, so if I want to define a state that is a superposition of the vacuum state, and the 1 particle state, how would I do so without running into problems with units? Naively, I would write

$$|\Psi\rangle=|0\rangle+a^{r\dagger}_{\mathbf{p}}|0\rangle$$​

but this would be adding two expressions with different units, so that's not right for sure.

 

[Meir Achuz]

Correction:

The connection between box and Dirac normalization involves
$$\sum_n\rightarrow Vol\int d^3p$$.
The Vol is usually set equal to 1 and its units forgotten about.
Its best just not to worry about units.

 

[TriTertButoxy]

But what if I want to worry about units?

 

[CarlB]

But what if I want to worry about units?

Your question delves deep into the mysteries of the meaning of the vacuum state. I think your worries are very well founded.

If you want to see what Julian Schwinger thought about the vacuum, splurge $20 on your physics education and pick up a copy of "Quantum Kinematics and Dynamics". I should note that this is the only book that has these heretical thoughts of his; a later book that also covers Schwinger's "measurement algebra" has had the discussion of the vacuum removed. It's a classic text, you only need to read the first few chapters, and it is very easy reading. It's considered one of the most elegant introductions to QFT every written.

The book is cheap, go buy it. It's a quite common book. If you're studying QFT at a university I would think that the local bookstore would have it if the library does not. I won't tell you the answer, you'll have to look it up for yourself. Maybe my interpretation of what Schwinger wrote is incorrect. Maybe not.

Schwinger, along with Feynman, got the Nobel prize for figuring out how to cancel infinities, but then became unhappy with the theory and tried to move on to something new. The mainstream was happy to "shut up and calculate", and moved physics forward for the next 50 years. But foundational issues have a way of coming back.

Now I need to tell you that what I am writing here is quite heretical from the point of view of the physics establishment. You are going to be told to shut up and calculate. This is good advice, if you can conquer your curiousity.

Ah, what the heck. Take a good look at a creation operator. Applied to the vacuum state, it creates a particle where there was none before. But this is something that never occurs in nature. Instead, any time something is created, something else is annihilated. So are you entirely certain that it makes sense to split these two events and to talk about creation and annihilation as if they could be thought of as separate, divisible things? I don't.

Carl

 

[masudr]

Carl do you have a problem with ladder operators for the quantum simple harmonic oscillator?

 

[TriTertButoxy]

The ladder operators for the quantum simple harmonic oscillator are unitless.

Those for the theory of quantum fields are not...
...this is the source of all my problems.

Argh!

 

[CarlB]

Carl do you have a problem with ladder operators for the quantum simple harmonic oscillator?

I don't have a problem with them. I'm simply noting that the states are elements of physical reality, while the mathematical tools we use for moving from one state to another are not. Instead, the mathematical tools are just methods used to solve mathematics problems.

You can't change the energy of a simple harmonic oscillator without changing the energy of something else. The same applies to the raising and lowering operators for spin. Like energy, total angular momentum is conserved.

I say any method that works is a good method. But if you're looking for the foundations of a theory, it is sometimes useful to be able to recognize what is a part of physics and what is a part of mathematics. This is true especially when your theory has been stuck for 30 years and you want to push it to the next step.

What we've done for 30 years is to develop the theory of elementary particles on the basis of an expansion of those features of the basic quantum theory that are not necessary for the physical description of states. Instead of concentrating on the states themselves, we have concentrated on how one gets from one state to another (mathematically).

We've had great success by postulating symmetries that tell how to get from one state to another. But one must remember that it is the physical states themselves that are physical, not the symmetries that one uses to get from one state to another.

Of course all this is just my opinion, but when string theory gets itself tangled up in 10^{500} vacuum states, and starts talking about anthropic principles, it should give you enough of a reason for reconsidering the foundations of the theory that led to it.

There is a way of analyzing physics without having to deal with unphysical state transitions. The central idea is that you must upgrade your states so that they become operators. In standard QM this is done by going from the state vector description of a state to the density operator description. You can then analyze the relationships of the states by analyzing the relationships of the operators.

What you lose is linearity. That is enough to prevent most physicists from exploring the consequences. But what you gain is the elimination of the things that depend on mathematical choice in the description of the state (also called gauge freedom) in favor of the things that correspond to the physics (the state itself).

Carl

 

[masudr]

Take a good look at a creation operator. Applied to the vacuum state, it creates a particle where there was none before. But this is something that never occurs in nature. Instead, any time something is created, something else is annihilated.

You then said

You can't change the energy of a simple harmonic oscillator without changing the energy of something else. The same applies to the raising and lowering operators for spin. Like energy, total angular momentum is conserved.

This was exactly the point I was making (and I'm glad you got it from my question). For an SHO you can't use the ladder operators without getting/giving up energy from/to somewhere. Same with the angular momentum ones. Same with those for the quantum field.

But the whole point of the SHO exercise is not to get the ground state and actually raise it up to the 3rd excited state, say, by 3 repeated applications; instead it is merely to solve the energy eigenvalue problem $H\psi=E\psi$ (with H being the SHO Hamiltonian) in a more elegant way than solving an ugly differential equation. In the end, we have a list of wavefunctions and energy eigenvalues, and that's all we set out to do. That's why we don't have to worry about where the energy comes from.

It even boils down something a lot simpler. Take the first problem you ever considered (probably): the particle in a box. When one solves for the eigenvalues and eigenfunctions, one doesn't worry about where the energy comes from to put it in the higher states; one notes that these are the possible energy eigenstates allowed by the Rules of Quantum Mechanics.

 

[TriTertButoxy]

And it works, because the SHO ladder operators are unitless, but those of QFT aren't...

what's the fix?

 

[masudr]

I refer you to:

http://en.wikipedia.org/wiki/Creation_and_annihilation_operators#The_vacuum_state

I think that answers your question (namely the comparison of the vacuum state and non-vacuum states).

(Please bear in mind that I have no formal teaching in QFT yet, so this is at best, a guess.) It might have to do with how the Fock space is constructed out of tensor products of other spaces, which might not explicitly define separate units.

Other than that, all I can do is hope someone else who has learned about QFT can comment (I'm still an undergraduate).

 

[CarlB]

Masudr, that was a great article on creation, annihilation and the vacuum state. Wikipedia has a number of good articles on physics. A lot of them are clearer than that found in the textbooks.

Carl

 

[TriTertButoxy]

Hmmm... I've already searched through Wikipedia articles, and the one to which you pointed me simply clarifies the convention used for the notation of the ground state and the zero state.

But thanks anyway!

What I'm looking for is a fix to the apparent mismatch of units between any pair of states with a different number of particles. Notice that this arises due to the fact that the ladder operators in QFT are unitfull. So a supperposition of two states with a different number of particles is meaningless, by dimensional analysis.

This seems to be a very big problem, since this is part of the procedure of writing down a QM state with macroscopic strength. There's no possibility that nobody has noticed this mismatch.

Any ideas?

 

[masudr]

When one writes

$$|\psi\rangle=|0\rangle + |1\rangle$$

one really means

$$|\psi\rangle=|0\rangle \otimes |\rangle + |\rangle \otimes |1\rangle$$

because the space that $|\psi\rangle$ is really a tensor product of the vacuum-state space, the one-particle state space, the two-particle state space etc. i.e. $|0\rangle$ and $|1\rangle$ don't live in the same space to start with, so it doesn't make any sense at all to add them together, as you have noted.

The point is $|0\rangle$ isn't really in different units from $|1\rangle$ since they are both actually tensor products of the same kind.

This was the point I was trying to make without going all-out and saying all that, 'cos, well, it could be incorrect.

 

[TriTertButoxy]

Awsome! Now things are making a little more sense. Thanks, masudr.

So how does the electromagnetic field operator, $\hat{A}_\mu$, act on $|\psi\rangle$?

 

[CarlB]

What I'm looking for is a fix to the apparent mismatch of units between any pair of states with a different number of particles. Notice that this arises due to the fact that the ladder operators in QFT are unitfull. So a supperposition of two states with a different number of particles is meaningless, by dimensional analysis.

There's an interesting violation of dimensional analysis in the branch of geometry called the Geometric Algebra. It's basically Clifford algebra attached to a spacetime manifold. The gamma matrices of the Dirac equation form an example of this.

The fundamental differential operator is the familiar

$$\gamma^\mu\; \partial_\mu$$

The odd behavior is what sorts of units to put on the gamma matrices themselves. In examining the above, it seems like they should have units of length or time, but this is incompatible with, for example:

$$\gamma^3\;\gamma^3 = 1$$

Carl

 

[masudr]

Shouldn't that be

$$\gamma^3 \gamma_3 =1?$$

 

[CarlB]

Ooops. Sorry, I'm using the "east coast" metric, (-+++), you're using the "west coast" metric, (+---). But the units comment goes either way.

Incidentally, the reason I prefer the east coast metric is that I prefer to have my spatial dimensions have positive signature. It makes the whole thing a little closer to a Euclidean metric. Once you choose one or the other, it's easy to forget that there is a whole world of physicists out there using the wrong signature.

The classic paper on the subject is:
http://www.arxiv.org/abs/gr-qc/9704048

It turns out that choice of signature doesn't matter for the standard model, but makes a difference in real Clifford algebras.

Carl

 

[TriTertButoxy]

When one writes

$$|\psi\rangle=|0\rangle + |1\rangle$$

one really means

$$|\psi\rangle=|0\rangle \otimes |\rangle + |\rangle \otimes |1\rangle$$

because the space that $|\psi\rangle$ is really a tensor product of the vacuum-state space, the one-particle state space, the two-particle state space etc. i.e. $|0\rangle$ and $|1\rangle$ don't live in the same space to start with, so it doesn't make any sense at all to add them together, as you have noted.

The point is $|0\rangle$ isn't really in different units from $|1\rangle$ since they are both actually tensor products of the same kind.

This was the point I was trying to make without going all-out and saying all that, 'cos, well, it could be incorrect.

I'm not really sure what

$$|\psi\rangle=|0\rangle \otimes |\rangle + |\rangle \otimes |1\rangle$$​

means. Also, how would I generalize this to a superposition of three states?
Also, can you clarify what $|\rangle$ means?

 

[samalkhaiat]

If you inspect the QED lagrangian, you'll find that the electromagnetic vector potential must have the units of [1] energy.

According to Peskin and Schroder's QFT text, the expansion of the electromagnetic field operator is

$$A_\mu(x)=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf{p}}}}\sum_{r=0}^3\bigg(a_{\mathbf{p}}^r\epsilon_\mu^r(p)e^{-ip\cdot x}+a_{\mathbf{p}}^{r\dagger}\epsilon_\mu^{r*}(p)e^{ip\cdot x}\bigg).$$​

3 factors of energy from $d^3p$, and -1/2 factor from $1/\sqrt{2E_{\mathbf{p}}}$. If the polarization vectors, $\epsilon_\mu^r$, are unitless, then the ladder operators must carry units of 1½.

There is no problem at all. It is always possible to choose normalization in such away that the ladder operators become dimensionless;

Method(i): The simple box norlalization; expand the field as

$$A^{\mu}(x)=\sum_{r,\vec{k}}(2 L^{3}E)^{-1/2}\epsilon_{r}^{\mu}(\vec{k}) a_{r,\vec{k}} e^{-ikx} + ...$$

where the k summation is over all momenta allowed by the periodic boundary conditions (discretized momenta);
$$\vec{k}=\frac{2\pi}{L}\vec{n}$$
where $n=0,\pm1,\pm2,...$.
Now, in the mass unit [E]=[A]=-[L]=1, we have;

$$[A]=-1/2[E]-3/2[L]+[a], \Rightarrow[a]=0$$.

Now, do whatever you want with these dimensionless ladder operators.
Check Mandl & Shaw, they do QFT in a box.
Method(ii); If you want to work with the (continuum) integral expansion of the field, then you need to know the difference between operator and operator-valued distribution. In QFT, the solution of the field equation is an operator-valued distribution. There is a mathematically well-founded procedure in which the field distribution can be smeared out with a "good" function in such away that it (the field) becomes a bona fide operator linear with respect to this good function.
In QFT the relation
$$[a(p),a^{\dagger}(q)]=\delta^{3}(p-q)$$
seems to imply that the field consists of a noncountable set of oscillators labelled by the 3-momentum. However, it can be shown that we may use a countable set of oscillators instead of a continuum. This is achieved by using a set of good functions;
$$f_{k(i)}(\vec{p})\in L^2, i=1,2,..$$,
satisfying
1) orthonormality

$$\int (d^3p) f_{k(i)}(p) f_{k(j)}^{*}(p)=\delta_{ij}$$

2) completeness

$$\sum_{i} f_{k(i)}(p) f_{k(i)}^{*}(q)=\delta^{3}(\vec{p}-\vec{q})$$

Notice that [f]=-3/2 mass unit.
Now, we introduce (in your expansion) the smeared out operators (no longer distributions) by;
$$a_{k(i),r}=\int d^3pf_{k(i)}(p)a_{r}(p)$$
$$a_{k(i),r}^{\dagger}= \int d^3pf_{k(i)}^{*}(p)a_{r}^{\dagger}(p)$$

You can see that these operators are dimensionless, because [a(p)]=-3/2 in your expansion.

Now, you can write your commutation relations in terms of these operators

$$[a_{k(i),r},a_{k(j),s}^{\dagger}]=-g_{rs}\delta_{ij}$$

and your(dimentionless) one photon state becomes

$$|1_{k,r}\rangle = \int d^3p f_{k}^{*}(p)a_{r}^{\dagger}(p)|0\rangle = a_{k,r}^{\dagger}|0\rangle$$,

and

$$\langle1_{k,r}|1_{k,r}\rangle = \langle0|a_{k,r}a_{k,r}^{\dagger}|0\rangle = -g_{rr}\langle0|0\rangle$$
(the norm of the state containing a scalar photon(r=0) is negative)

And finally, your number operator becomes
$$N=\sum_{k(i),r}a_{k(i),r}^{\dagger}a_{k(i),r} = \sum_{r} \int d^3p(-g_{rr})a_{r}^{\dagger}(\vec{p})a_{r}(\vec{p})$$
This is dimensionless as it should be.

good luck

regards

sam