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  Why, in spin sums, we sum over final spin states and average over initial states?

+ 5 like - 0 dislike

I am reading Halzen's book about quarks and leptons and on page 120 he talks about spin sums.

He says that in order to calculate the amplitude between unpolarized states we have to sum over FINAL spin states and average over INITIAL states. Why is this so? why not sum over initial states and average over finals, or sum over initial and final spin states and average also over initial and final states?

This post imported from StackExchange Physics at 2015-03-23 06:04 (UTC), posted by SE-user silvrfück
asked Sep 25, 2014 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ no revision ]
retagged Mar 23, 2015

3 Answers

+ 4 like - 0 dislike

For the incoming state, you don't know which spin state the particle is in, so you should average over the possible states. But you can measure the spin of the outgoing state, so to get the total cross section you should add up the cross sections for each spin.

More formally, an unpolarized incoming particle should be described as a density matrix, $$ \def\bra#1{\mathinner{\langle{#1}|}} \def\ket#1{\mathinner{|{#1}\rangle}} \rho(t=-\infty) = \frac{1}{2}( \ket{p\uparrow} \bra{p\uparrow} + \ket{p\downarrow}\bra{p\downarrow})$$ where the $\frac{1}{2}$ is for the normalization of the density matrix, $\operatorname{tr} \rho = 1$. Here $p$ is the momentum and this is of course for spin $\frac{1}{2}$ but the generalization to spin $1$ and higher will be obvious.

The density matrix will evolve to $\rho(t=\infty) = S \rho(t=-\infty) S^\dagger$ just by definition of the $S$-matrix. You want to calculate the probability to end up with momentum $q$, regardless of spin, at $t=\infty$. This is the expectation value at $t=\infty$ of the projection operator $$ P(q) = P(q,\uparrow) + P(q,\downarrow) = \ket{q\uparrow}\bra{q\uparrow} + \ket{q\downarrow}\bra{q\downarrow}. $$

The expectation value of the projection operator is $$\langle P(q) \rangle = \operatorname{tr}(\rho P(q) ).$$ There will be four terms in the trace. One of them is $$T_{\uparrow\uparrow} = \frac{1}{2}\operatorname{tr}\big(S\ket{p\uparrow}\bra{p\uparrow}S^\dagger\ket{q\uparrow}\bra{q\uparrow}\big).$$ Note that in the middle is something that is just a number, and a familiar number too. $\bra{p\uparrow}S^\dagger\ket{q\uparrow} = (\bra{q\uparrow}S\ket{p\uparrow})^*$ is (the complex conjugate) of an S-matrix element. It is easy to realize that $$\operatorname{tr}\big(S\ket{p\uparrow}\bra{q\uparrow}\big) = \bra{q\uparrow}S\ket{p\uparrow}$$ and therefore $$T_{\uparrow\uparrow} = \frac{1}{2} |\bra{q\uparrow}S\ket{p\uparrow}|^2.$$ The other terms in the trace are similar and we end up with $$\langle P(q) \rangle = \frac{1}{2} \sum_{r,s} |S(p,r;q,s)|^2$$ where $r,s \in \{\uparrow, \downarrow\}$ are incoming and outgoing spins, respectively, and $S(p,r;q,s)$ is the $S$-matrix element. As you can see we recover the prescription average over incoming, and sum over outgoing.

This post imported from StackExchange Physics at 2015-03-23 06:04 (UTC), posted by SE-user Robin Ekman
answered Sep 25, 2014 by Robin Ekman (215 points) [ no revision ]
+ 3 like - 0 dislike

Imagine an experiment where you are sending one particle of spin 1/2 into a "black box" and waiting for the result.

You know you are sending in one particle at a time, but you do not know the spin and presume it's going to be fifty fifty - so you say in fifty percent of cases you will have the spin "up" (or better "right" for relativistic particles) and add the scenario with a fifty percent weight $50\%=50 \frac{1}{100}=\frac{1}{2}$ - you do the same for the opposite spin "down" ("left"). I.e. you average over the incoming particles (remember we are always talking about multiple repetitions of the experiment where different particles are sent in during every repetition)

Now imagine the black box spits out two particles. They may have spin up+spin up in 50% of cases, or spin down + spin down in the other 40% of cases and no particle flies out in 10% of cases. But if you don't care about the spin at all and want to know in how many cases you get two outcoming particles, you just sum these scenarios without any averaging. It's common sense you just add the probabilities (aka "amplitudes") of all events of two outcoming particles and get a probability of $90\%$.

It's just common sense and probability, the only part where quantum mechanics enters is the inner workings of the black box and the fact that the results are probabilistic.

This post imported from StackExchange Physics at 2015-03-23 06:04 (UTC), posted by SE-user Void
answered Sep 25, 2014 by Void (1,645 points) [ no revision ]
I believe the OP is talking about adding amplitudes, not probabilities.

This post imported from StackExchange Physics at 2015-03-23 06:04 (UTC), posted by SE-user Peter Shor
But spin sums are done over the quadrates of moduli of matrix elements, which is exactly proportional to probabilities of outcomes.

This post imported from StackExchange Physics at 2015-03-23 06:04 (UTC), posted by SE-user Void
+ 2 like - 0 dislike

This is simply a consequence of basic properties of probabilities.

Let us take an event $A$ made of independent events $A_i$ (so that $p(A) = \sum\limits_i p(A_i)$), and an event $B$ made of independent events $B_j$ (so that $p(B) = \sum\limits_j p(B_j)$)

Then you have : $p(A/B) = \dfrac {\sum\limits_{i,j} p(A_i/B_j) \,p(B_j)}{\sum\limits_{j}\,p(B_j)}$

This post imported from StackExchange Physics at 2015-03-23 06:04 (UTC), posted by SE-user Trimok
answered Sep 26, 2014 by Trimok (955 points) [ no revision ]

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