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In quantum field theory, the elements of the S-matrix are defined as the amplitude describing the transition from an initial $n$-particle state (the "in" state) to an final $m$-particle state: \begin{equation} S_{fi} = \langle \mathbf{q}_1,\dots,\mathbf{q}_m; \text{out} | \mathbf{p}_1,\dots,\mathbf{p}_n ; \text{in} \rangle \tag{1} \end{equation} To me it seems that this equation only makes sense if the amount of "in" particles is equal to the amount of "out" particles (i.e. $m=n$) otherwise we can not take the inner product. For instance, if $m=2$ and $n=3$, then we can write equation $(1)$ as: \begin{equation} \begin{pmatrix} \mathbf{q}_1 & \mathbf{q}_2 \end{pmatrix} \begin{pmatrix} \mathbf{p}_1 \\ \mathbf{p}_2 \\ \mathbf{p}_3 \end{pmatrix} =\ ? \end{equation} which is undefined. Therefore, my question is how to interpret equation $(1)$ if $m \neq n$?

I must admit that I have never studied the S-matrix in quantum mechanics. Therefore, I apologize in advance if this is a naive question.

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Your second equation is a misinterpretation of the notation in your first equation. $|q_1,q_2\rangle$ is not a vector in a finite dimensional vector space, having components $q_1$ and $q_2$ in some basis. Rather $|q_1,q_2\rangle $ and $|p_1,p_2,p_3\rangle$ are both vectors in the same infinite-dimensional Hilbert space.

In the simplest case, where all particles are bosons of the same species, this Hilbert space is spanned by the following collection of 'in' vectors.

  1. $|0\rangle$ -- a single vector of unit norm, representing the state with no incoming particles.
  2. $|p\rangle$ -- a unit vector for each momentum $p$. If $p'$ and $p'$ are different momenta, then $|p\rangle \neq |p'\rangle$. These vectors are states where we have precisely one particle coming in from past infinity with a fixed momentum.
  3. $|p_1,p_2\rangle$ -- a unit vector for each pair $(p_1,p_2)$ of momenta. $|p_1,p_2\rangle$ is not $|p_1',p_2'\rangle$ unless $p_1 = p_1'$ and $p_2=p_2'$ or $p_1 = p_2'$ and $p_2=p_1'$. These are states where we have two particles coming in from past infinity.
  4. $|p_1,p_2,p_3\rangle$ -- a unit vector for each triplet $(p_1,p_2,p_3)$. These vectors are only equal to each other if their labels $(p_1,p_2,p_3)$ are permutations of each other.
  5. ...

This is a continuous basis, which can seem a little weird if you're used to finite dimensional vector spaces. Finite sums get replaced by sum over number of particles and integrals over the momentum labels.

There's a similar basis $|0\rangle, \{|q\rangle\}, \{|q_1,q_2\rangle\},..$ of 'out' vectors, where now our momentum labels states according to the number of particles at future infinity with fixed momenta.

The S-matrix is the change of basis coefficients for switching from the in-basis to the out-basis.

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    $\begingroup$ Thanks, this is kind of the answer that is most on my level of current understanding and is exactly what I needed! $\endgroup$ – Hunter Jan 18 '14 at 22:00
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    $\begingroup$ I should also comment that many authors use a convention in which the vectors aren't normalized to have length 1. This is done purely for convenience; it has no physical content, since you can represent a state using any non-zero vector in the ray. $\endgroup$ – user1504 Jan 18 '14 at 22:01
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    $\begingroup$ It may aid your learning to think about how @V.Moretti's answer generalizes mine. (For example: Imagine a system where particles can be of either of 2 species. Now you need to keep track of whether the momentum is for a particle of type 1 or type 2.) $\endgroup$ – user1504 Jan 18 '14 at 22:06
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The Hilbert space of the theory $\cal H$ can be viewed simultaneously as a Fock space in two different manners: $${\cal H} = {\cal F}_{symm}(K_{in}) = {\cal F}_{symm}(K_{out})\:.$$ Where $K_{in}$ and $K_{out}$ are the one-particle space of ingoing and outgoing free particles and I am assuming that the particles are Bosons for the sake of simplicity. Above: $${\cal F}_{symm}(K) = {\mathbb C}\oplus K \oplus (K\otimes K)_{symm} \oplus (K\otimes K\otimes K)_{symm}\oplus \cdots $$

The point is now that, in general $K_{in}$ include not only vectors of $K_{out}$, but even of $(K_{out}\otimes K_{out})_{symm}$, $(K_{out}\otimes K_{out}\otimes K_{out})_{symm}$, and so on. In other words it is false that, for instance $K_{in} \not\perp (K_{out}\otimes \cdots (k \:times)\cdots \otimes K_{out})_{symm}$ for $k>1$.

This is the mathematical translation of the fact that an ingoing free particle, in the asymptotic past (when the interactions are switched off), due to interactions at finite time, may give rise to many free particles in the asymptotic future (when the interactions are again switched off). In general the number of particles is not preserved due to the interactions passing from $t=-\infty$ to $t=+\infty$.

In general, for every value of $n$ and $m$:

$(K_{in}\otimes \cdots (n \:times)\cdots \otimes K_{in})_{symm} \not\perp (K_{out}\otimes \cdots (m \:times)\cdots \otimes K_{out})_{symm}$.

These relations can be written using vectors:

$$\langle\psi^{(out)}_{1}\cdots \psi^{(out)}_{m}| \psi^{(in)}_{1}\cdots \psi^{(in)}_{n}\rangle \neq 0 \quad \mbox{for generic $n\neq m$,}$$

where, for instance

$$ |\psi^{(out)}_{1}\cdots \psi^{(out)}_{m}\rangle \in (K_{out}\otimes \cdots (m \:times)\cdots \otimes K_{out})_{symm}$$

is the generic symmetrised state made of $m$ outgoing particles with single states $\psi^{(out)}_{1},\cdots, \psi^{(out)}_{m}\in K_{out}$, not necessarily pairwise different.

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  • $\begingroup$ Thanks for your answer and +1. I should really one day buy "PCT, Spin and Statistics, and All That" (as you have recommended me in another post), because one day I hope to understand this as well as you do. $\endgroup$ – Hunter Jan 18 '14 at 21:59
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    $\begingroup$ I became a mathematician but I had beforehand learned these things using "hands and feet" as a physicist. I think that is the correct way to understand QFT. Later, one can pass to more formal books "PCT, Spin and Statistics, and All That" or Haag's textbook...so, do not worry you are doing the right thing, for the moment at least. $\endgroup$ – Valter Moretti Jan 18 '14 at 22:05
  • $\begingroup$ @V.Moretti I couldn't agree more. Also, to which textbook of Haag's are you referring? $\endgroup$ – joshphysics Jan 18 '14 at 22:23
  • $\begingroup$ Local Quantum Physics $\endgroup$ – Valter Moretti Jan 18 '14 at 22:31
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What you seem to be thinking of is the "leading" comtribution to the amplitude -- where the each in-particle goes to an out-particle without interacting. This happens in any free/linear theory of waves passing through each other without interacting.

What we're interested in is the $T$ matrix in $S = \mathbf{1} + i T$ while you're talking about contribution from the $\mathbf{1}$. Imagine a simple "interaction" where one particle decays to a pair of particles. Such an aspect of the theory (among other possibilities) will contribute to the $T$ matrix and give you non-zero amplitudes even when the number of particles changes (maybe increases by one).

To put it another way, there is no reason that the number of particles ought to be conserved in a process, except in maybe special cases. Can you think of any symmetry that might imply such a conserved quantity?


Mathematically, in the perspective of states in a Hilbert space:

Without getting into deep questions about the "existence" of a Hilbert space for an interacting theory (which I don't know enough to comment on)... For an interacting theory, the space of states is the Fock space $\mathcal{H} = \oplus H_n$ where $H_n$ is the n-particle Hilbert space. Each of those states can be denoted by their quantum numbers under symmetries of the theory (representation theory).

$\langle q_1, q_2 | p_1, p_2 ,p_3 \rangle$ is NOT like $(q_1, q_2) \cdot (p_1, p_2, p_3)$. You should think of it as an inner product of some two states $\langle \Psi_2 | \Psi_3 \rangle$. $|\Psi_2 \rangle$ happens to be a certain state in the Fock space, parametrized by the two quantum numbers $q_1 , q_2$ (similar interpretation for $|\Psi_3 \rangle$). Each of those are states in an infinite dimensional vector/Hilbert/Fock space whose degrees of freedom can be thought of as excitations at each spatial point, for each field in the theory.

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  • $\begingroup$ From a physical point of view, I completely understand that the particles can change when they interact with each other (as has been determined by many experiments). My questions is more related to the mathematics of the Hilbert space and the inner product. $\endgroup$ – Hunter Jan 18 '14 at 19:59
  • $\begingroup$ Without getting into deep questions about the "existence" of a Hilbert space for an interacting theory (which I don't know enough to comment on)... For an interacting theory, the space of states is the Fock space $\mathcal{H} = \oplus_n H_n$ where $H_n$ is the n-particle Hilbert space. Does that help? $\endgroup$ – Siva Jan 18 '14 at 21:07
  • $\begingroup$ Hmmm, I am not sure. The Fock space I believe is spanned by: $$\{ |\mathbf{p}_1\rangle, |\mathbf{p}_1,\mathbf{p}_2\rangle,|\mathbf{p}_1,\mathbf{p}_2,\mathbf{p}_3\rangle, \ldots \}$$Therefore, it makes sense to think that it is somehow useful for the interacting space. I am now wondering if I am thinking my question is unnecessarily complicated and I should just accept the way the S-matrix is defined (especially because none of the sources, such as Pesking & Schroeder, even mention this problem I am having). $\endgroup$ – Hunter Jan 18 '14 at 21:17
  • $\begingroup$ $\langle q_1, q_2 | p_1, p_2, p_3 \rangle$ is NOT like $(q_1 q_2) \cdot (p_1, p_2, p_3)$. You should think of it as an inner product of some two states $\langle \Psi_2 | \Psi_3 \rangle$. $| \Psi_2 \rangle$ happens to be a certain state in the Fock space, parametrized by the two quantum numbers $q_1, q_2$ (similar interpretation for $|\Psi_3 \rangle$). Each of those are states in an infinite dimensional vector/Hilbert/Fock space whose degrees of freedom can be thought of as excitations at each spatial point, for each field in the theory. $\endgroup$ – Siva Jan 18 '14 at 21:30
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    $\begingroup$ Sure; I'll gather stuff from my comments and move it to the "answer". $\endgroup$ – Siva Jan 18 '14 at 21:36

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