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When adding the angular momenta of two particles, you use Clebsch-Gordan coefficients, which allow you, in fancy language, to decompose the tensor product of two irreducible representations of the rotation group into a direct sum of irreducible representations. (I'm not exactly clear on what this means, so on a side note can someone suggest a good book on representation theory of Lie groups?)

If we want to add the angular momenta of even more particles, then we have to use other coefficients, like the Wigner 3j, 6j, or in general 3nj symbol. But what if you have an unknown number of particles, like you do in quantum field theory? In this case quantum states live in a so-called Fock space made of a direct sum of infinitely many tensor products of Hilbert spaces (which is another thing I'm unclear on). So then how do you add angular momentum in Fock space? To put it another way, how do you decompose a tensor product of irreducible representations into a direct sum if you have a variable number of terms in the tensor product?

Any help would be greatly appreciated. Thank You in Advance.

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So it sounds like if you knew how many particles there were, you would be comfortable with the calculation. Then the Fock space answer isn't really any more complicated... The goal is to write the whole space as a direct sum of irreducible parts. Since the Fock space is a direct sum of hilbert spaces with definite particle number, you can then just split each hilbert space into a direct sum of irreps of the rotation group, and you are done. Actually it's a bit more complicated because you really need reps of the poincaire group, not the rotation group, but let's say everything is at rest. –  Andrew Oct 12 '13 at 15:36
    
Yes, I just want to use the rotation group. But can you (or someone else) write out what decomposition would look like? I'm a bit shaky as far as tensor products and irreps go. –  Keshav Srinivasan Oct 12 '13 at 15:41
    
Side note: I'd recommend two books, namely Hall for general representation theory and De Melo for its applications to QFT. –  Edward Hughes Oct 12 '13 at 18:52
    
Another comment. Andrew's comment shows correctly that it's possible to do what you ask. The question of the exact coefficients would be very messy though. You'd essentially get higher particle number analogues of the Clebsch Gordan coefficients. I wouldn't be surprised if you could just calculate these recursively somehow. In my experience it's not important for QFT calculations though. –  Edward Hughes Oct 12 '13 at 18:59
    
@EdwardHughes Higher-particle number analogues of Clebsch-Gordan coefficients have already been found. Look up Wigner 3j symbol, Wigner 6j symbol, and more generally Wigner 3nj symbol. –  Keshav Srinivasan Oct 12 '13 at 19:48
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First of all, I got a bit carried away writing this response, so it's very long (but hopefully useful); my apologies for any grammar errors

As a book reference for introductory representation theory, I personally like the same book that Edward Hughes recommended.

Lie Groups, Lie Algebras, and Representations by Brian Hall.

It's a math book written by a mathematician, but it's not overly abstract for a physicist, and if you want to learn representation theory right, your best bet is probably to let a mathematician teach it to you. (I hope I don't get hate mail for saying that.)

As for the Fock space question, I think one needs to be careful to distinguish between (a) writing down an infinite direct sum of tensors products of spin Hilbert spaces and performing Clebsch-Gordan decomposition and (b) writing down a physically relevant Fock space, and attempting to decompose its direct summands. Performing (a) doesn't involve any real subtleties as indicated by Andrew and Edward Hughes, and here's how you would do it explicitly. Suppose that we have an $n$-fold tensor product of angular momentum Hilbert spaces; \begin{align} \mathcal H_{j_1}\otimes\mathcal H_{j_2} \otimes \cdots \otimes \mathcal H_{j_n}, \end{align} then we can perform a Clebsch-Gordan decomposition recursively in pairs by noting that the tensor product distributes over the direct sum in much the same way that multiplication of real numbers distributes over addition, and by taking advantage of the associativity of the tensor product. We could start with the first pair $\mathcal H_{j_1}\otimes\mathcal H_{j_2}$ which gives \begin{align} \mathcal H_{j_1}\otimes\mathcal H_{j_2} = \bigoplus_{j=|j_1-j_2|}^{j_1+j_2} \mathcal H_j \end{align} We can then deal with the first three factors; \begin{align} \mathcal H_{j_1}\otimes\mathcal H_{j_2}\otimes\mathcal H_{j_3} = \bigoplus_{j=|j_1-j_2|}^{j_1+j_2} \mathcal H_j\otimes\mathcal H_{j_3} = \bigoplus_{j=|j_1-j_2|}^{j_1+j_2}\bigoplus_{j'=|j-j_3|}^{j+j_3} \mathcal H_{j'} \end{align} and so-on. So in the event that you consider some infinite direct sum like \begin{align} \mathbb C \oplus \mathcal H_{j_{11}} \oplus (\mathcal H_{j_{21}}\otimes\mathcal H_{j_{22}}) \oplus (\mathcal H_{j_{31}}\otimes\mathcal H_{j_{32}}\otimes\mathcal H_{j_{33}})\oplus\cdots, \end{align} one can apply the above procedure to each direct summand to eliminate all tensor products in favor of direct sums.

But now, what about case (b) where we want to consider some the Fock space of some physical system. In this context, Fock space usually refers to an infinite direct sum of symmetrized or antisymmetrized tensor products of a certain single-particle Hlbert space with itself. Physically, this comes from the fact that it's the sort of space we use to describe systems consisting of an arbitrary number of identical fermions of bosons. Since we have to be careful about particle symmetrization and antisymmetrization in each direct summand, the story is rather subtle.

Let's consider, for the sake of concreteness, a system that can contain an arbitrary number of identical spin-$1/2$ particles. Let $\mathcal H_\frac{1}{2}$ denote the spin-$1/2$ Hilbert space. Then we might naively think that the Fock space for such a system is as follows: \begin{align} \mathbb C \oplus S_-(\mathcal H_\frac{1}{2}) \oplus S_-(\mathcal H_\frac{1}{2}\otimes \mathcal H_\frac{1}{2}) \oplus S_-(\mathcal H_\frac{1}{2}\otimes\mathcal H_\frac{1}{2}\otimes\mathcal H_\frac{1}{2})\oplus\cdots \end{align} The first term $\mathbb C$ in the direct sum corresponds to the Hilbert space with no particles in it, which is one-dimensional. The symbol $S_-$ in front of each summand picks out the antisymmetric subspace of the Hilbert space on which it operates since we are dealing with identical fermions whose states must be totally antisymmetric. Now the question becomes, how does one decompose this beast into direct sums of Hilbert spaces, each of which is a copy of some spin-$s$ Hilbert space? Let's deal with the direct summands one-by-one. First, we note that \begin{align} S_-(\mathcal H_\frac{1}{2}) = \mathcal H_\frac{1}{2} \end{align} since when there is a single particle, no antisymmetrization does nothing; there are no tensor factors to permute. Next, we use Clebsch-Gordan theory for the second summand; \begin{align} \mathcal H_\frac{1}{2}\otimes \mathcal H_\frac{1}{2} &\cong \mathcal H_0 \oplus \mathcal H_1 \end{align} where $\cong$ here denotes isomorphism of vector spaces. In other words, the spin-$1/2$ Hilbert space decomposes into a direct sum of a spin-$0$ and a spin-$1$ Hilbert space. Now, it is not hard to see by inspection that all of the (three) states in the spin-$1$ Hilbert space are symmetric, while the single state in the spin-$0$ Hilbert space is antisymmetric. If you don't believe me, open a book on quantum and stare at the "singlet" and "triplet" states which form bases for $\mathcal H_0$ and $\mathcal H_1$ respectively, and you'll see that the triplet states don't change under interchange of the two particles, but the singlet state changes by a sign. It follows that \begin{align} S_-(\mathcal H_\frac{1}{2}\otimes \mathcal H_\frac{1}{2}) \cong \mathcal H_0. \end{align} Now what about the terms with higher tensor powers of $\mathcal H_\frac{1}{2}$? If we have three factors of $\mathcal H_\frac{1}{2}$, then we can write \begin{align} \mathcal H_\frac{1}{2}\otimes \mathcal H_\frac{1}{2} \otimes \mathcal H_\frac{1}{2} &\cong (\mathcal H_0\oplus \mathcal H_1)\otimes \mathcal H_\frac{1}{2} \\ &\cong (\mathcal H_0\otimes \mathcal H_\frac{1}{2})\oplus (\mathcal H_1\otimes \mathcal H_\frac{1}{2})\\ &\cong (\mathcal H_\frac{1}{2})\oplus (\mathcal H_\frac{1}{2}\oplus\mathcal H_\frac{3}{2}) \end{align} But now, notice the following. A basis for the spin-$1/2$ Hilbert space consists of the spin up and spin down states $|+\rangle$ and $|-\rangle$. It follows that a general state $|\psi\rangle$ in this $3$-particle Hilbert space will some linear combination of tensor product basis elements of the form $|\nu_1\rangle|\nu_2\rangle|\nu_3\rangle$ where each $\nu_i = \pm$; \begin{align} |\psi\rangle = \sum_{\nu_1,\nu_2,\nu_3=\pm}c_{\nu_1,\nu_2,\nu_3}|\nu_1\rangle|\nu_2\rangle|\nu_3\rangle \end{align} But there is no such state that is totally antisymmetric under the exchange of any two tensor factors, so there is no state in this Hilbert space that is totally antisymmetric. This is just a reflection of the Pauli Exclusion principle; when the single-particle Hilbert space is only two-dimensional, one can only build composite systems out of this Hilbert space that contain at most two particles. In fact, similar reasoning shows that one cannot build a totally antisymmetric state in any of the Hilbert spaces that contain three or more tensor factors of $\mathcal H_\frac{1}{2}$. It follows that symmetrization of the $n$-fold tensor product of $\mathcal H_\frac{1}{2}$ with itself gives the zero vectors space for all $n\geq 3$; \begin{align} S_-(\mathcal H_\frac{1}{2}^{\otimes n}) = \{0\}, \qquad\text{for all $n\geq 3$} \end{align} and therefore, putting this all together shows that the Fock space actually just reduces to a finite-dimensional vector space; \begin{align} \mathbb C\oplus \mathcal H_\frac{1}{2}\oplus(\mathcal H_\frac{1}{2}\otimes \mathcal H_\frac{1}{2}) = \mathbb C \oplus \mathcal H_\frac{1}{2}\oplus \mathcal H_0 \end{align} But now you may ask, how then do we have systems that can contain, for example, an infinite number of identical electrons? Take, for example, a gas of non-interacting electrons in one dimension in the harmonic oscillator potential in contact with an infinite particle reservoir subject, in addition, to an external magnetic field in the $z$-direction to which the spin of each electron couples. In this case, the single-particle Hilbert space is no longer just $\mathcal H_\frac{1}{2}$, which is two-dimensional. Instead, the single-particle Hilbert space is \begin{align} \mathcal H = \mathcal H_\frac{1}{2}\otimes \mathcal H_\mathrm{ho} \end{align} where $\mathcal H_\mathrm{ho}$ denotes the Harmonic oscillator Hilbert space that is spanned by the states $|n\rangle$ with $n=0,1,2\dots$. In particular, the single particle Hilbert space is now infinite-dimensional. For notational convenience, let's write the tensor product basis elements of $\mathcal H$ as $|\nu, n\rangle$ where $\nu =\pm$ is the spin up-down label and $n$ is the harmonic oscillator label. In this case, the Fock space is \begin{align} \mathbb C\oplus S_-(\mathcal H)\oplus S_-(\mathcal H\otimes\mathcal H)\oplus S_-(\mathcal H\otimes\mathcal H\otimes\mathcal H)\oplus \cdots \end{align} But now, the the story is a lot richer than in the last example. Let's look, for example, at the third direct summand. In the last example where the single particle Hilbert space was just $\mathcal H_\frac{1}{2}$, the triplet states were not allowed in the double-particle Hilbert spaces because they were symmetric, and we are dealing with identical Fermions. But now that the single particle space has a Harmonic oscillator factor, the story is much different. In particular, it's easy to construct an antisymmetric state in which both of the particles are spin-up as follows: \begin{align} |+,0\rangle|+,1\rangle - |+,1\rangle|+,0\rangle \end{align} They key point here is that since there is now an extra tensor factor in the single-particle Hilbert space, there is another label, in this case a non-negative integer, and can be used to break the symmetry between the spin states of the two particles as in the state above where they are both spin up, but have different harmonic oscillator labels. This means that the two particle Hilbert space no-longer degenerates into just the singlet Hilbert space. In this case, in fact, it is infinite dimensional contains states of both spin $0$ and spin $1$, as in the full Clebsch-Gordan decomposition. Similar comments apply to the later direct summands. For example, in the fourth term where we have three tensor factors of the single-particle Hilbert space, we can have states with any combination of the three spins being up or down. For example, if we want all of the spins to be down then we can form the state \begin{align} \sum_{\sigma\in S_3}(-1)^\mathrm{sgn(\sigma)}|-,\sigma(0)\rangle|-,\sigma(1)\rangle|-,\sigma(2)\rangle \end{align} where $S_3$ is the group of permutations of $\{0,1,2\}$ and $\mathrm{sgn}(\sigma)$ is the sign of the permutation which is positive if it involves an even number of exchanges, and negative otherwise.

The main point, is that now, even after antisymmetrization, the antisymmetrization of each direction summand containing $n$ particles, namely \begin{align} S_-(\mathcal H^{\otimes n}) \end{align} contains states of every total spin that would have appeared in the Clebsch-Gorban decomposition of the $n$ spin space tensor factors alone. However, it's not clear to me if there is a systematic notation for writing down what the resulting antisymmetrized Hilbert space is, I'm not sure if the Clebsch-Gordan decomposition does much for us in these cases, and I've never personally seen this issue crop up in field theory/QM courses/research.

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A good book: K. Blum - Density Matrix Theory and Applications (Physics of Atoms and Molecules).

Now about your question, I'm gonna answer from my experience, since I work with particle physics and atomic physics together. I hope my answers are right. If I'm wrong, please correct me.

You CANNOT work a system without knowing the exact number of particles if you're going to work it particle by particle, however, no one ever does that. For example, we work with Cs magnetometers, which (basically) use a polarized laser beam to pump the angular momentum of Cs atoms, that meanwhile precess according to Larmor's law. The system would have the Avogadro number of particles, for example. Would it make sense to solve a system of a system differential equation for all those (density matrices $\otimes $ number of particles). Actually what we do is that we solve the differential equation for a mixture with only a single density matrix that takes care of everything. We never care about the number of particles. An example can be given in this paper:

http://arxiv.org/abs/physics/0605234

In other words, you always take approximations. No sane person would try to solve a 10^23 set of differential equations (while you always have to know the number of particles). We knowing how to do it doesn't mean we can do it.

Why can we do that? Because when you have a huge number of particles, the angular momentum starts to behave classically. So if you have a single electronic spin, you have states -1/2 and 1/2; if you have two, you have states -1,0 and 1. If you have 10, you'll have from -5,...,5 ... if you have a million: you'll have almost a continuous spectrum of spins.

So the short answer is: Always do approximations.

Now in Quantum Field Theory, you also have to know the number of particles, but in a different way. Though when you work with annihilation and creation of particles, the spin is always conservered! And actually if you refer to Feynman integrals, you'll notice that you have the very wide range of all possible outcomes of a field(s) (particle(s)), and then you integrate over all possible states. So you basically integrate over every possible result that could come out, and that's how you calculate the probability of the event happenening, and you should know what you're doing with how many particles, also.

And don't get me started on the approximations they use in the computations of Quantum Field Theory. You can solve the Cs atom problem in full detail with high accuracy because of the shell nature of the Alkali atoms that mimic the Hydrogen atom. While in Quantum Field Theory you have a gazillion terms that have to be integrated over with perturbation theory, which take ages to be calculated.

Hope this helps!

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I don't think "in Quantum Field Theory, you also have to know the number of particles" is right at all. Are you familiar with Fock space? And yes, there is conservation of angular momentum, but that doesn't change the complications that go into angular momentum decomposition over a fixed number of particles. So I'd like to know what the decomposition looks like if you have amplitudes for having different numbers of particles. –  Keshav Srinivasan Oct 12 '13 at 15:08
    
@KeshavSrinivasan I said "in a different way"; not exactly. What you have to know is that the total energy that converts from one field type to another and some other parameters. If the number of particles changes, then you have to redo the angular momentum analysis that you did previously for each different case you want to study. I feel from your wording that you want something like a magical recipe to follow, but such a thing doesn't exist. You start your analysis with the particles you have at hand, and then you analyze your angular momenta starting from there. –  The Quantum Physicist Oct 12 '13 at 15:19
    
Well, I basically want to see a formal calculation done in Fock space where we use the amplitude of having each number of particles and the appropriate Wigner 3nj symbol, and see how the tensor products of angular momentum states in different summands of the Fock Space add up. Like would you be able to combine terms from the expressions for different numbers of particles, etc. –  Keshav Srinivasan Oct 12 '13 at 15:26
    
@KeshavSrinivasan I think such a calculation is a research topic and isn't formally available in some text book. Most of Quantum Field Theory calculations are done with simulations, and not analytically. I don't even know a single QFT calculation that can be done analytically. Sorry to disappoint you! –  The Quantum Physicist Oct 12 '13 at 19:14
    
I don't want like an actual numerical value. I just want to see what the formal expression looks like. –  Keshav Srinivasan Oct 12 '13 at 19:47
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