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S-Matrix definition and terminology

  1. May 10, 2014 #1
    A lot of textbooks give the definition of an S-matrix element as:

    [tex]\langle \beta_{out}| \alpha_{in}\rangle = \langle \beta_{in}| S| \alpha_{in} \rangle=\langle \beta_{out}| S| \alpha_{out} \rangle=S_{\beta \alpha}[/tex]

    and that [itex]S|\alpha_{out} \rangle =|\alpha_{in} \rangle [/itex]

    I don't understand that definition. Shouldn't the S-matrix take an in-state, and map it to the corresponding out-state:

    [tex]S|\alpha_{in} \rangle=|\alpha_{out} \rangle [/tex]

    Moreover, shouldn't the amplitude for a state prepared in [itex]\alpha [/itex] to be detected as β be:

    [tex]\langle \beta_{out} | S|\alpha_{in} \rangle [/tex]?
     
  2. jcsd
  3. May 10, 2014 #2
    I don't know all the conventions but the definition I have always found is the first one:
    $$
    \psi^{in}_\alpha=\int d\beta S_{\beta \alpha}\psi^{out}_\beta.
    $$

    It depends on what formalism you are using. You can either construct the theory using only in states or you can built a formalism with out states as well. In this second case the scattering matrix is just given by:
    $$S_{\alpha\beta}=\langle \psi^{out}_\alpha|\psi^{in}_\beta\rangle. $$
    In a certain sense, all the information about the scattering is automatically embedded in the definition of in and out states. If you, instead, start with just in states then you need the scattering matrix (which is nothing but the Scrodinger evolution operator for very large times) to evolve from your in state to the final state (which is not an out state).

    You can find a very good description of these formalism in Weinberg's book "Lectures on Quantum Mechanics".
     
  4. May 10, 2014 #3
    To me it makes more sense to write your expression like this:

    $$
    S\psi^{in}_\alpha=\int d\beta S_{\beta \alpha}\psi^{out}_\beta.
    $$

    in analogy with how an operator acts O acts on a vector:

    $$
    O|a \rangle= \int db \, |b\rangle \langle b|O|a \rangle= \int db \, O_{ba} |b\rangle
    $$

    In the example above, O would be the scattering matrix S. S would them map an asymptotic in-state [itex]|a \rangle [/itex] to an asymptotic out-state [itex] \int db \, O_{ba} |b\rangle [/itex] which belong to the same Hilbert space.
     
  5. May 11, 2014 #4
    I think this is just a definition, and it can describe the system for sure. Just not by the usual way that make us feel comfortable. that's it.
     
  6. May 11, 2014 #5

    ChrisVer

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    Gold Member

    Would it play a role whether S maps in to out or out to in?
     
  7. May 11, 2014 #6

    ChrisVer

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    Gold Member

    For any case, the nicest definition of S-matrix I've seen is the time evolution one...
    [itex] <A_{out}|B_{in}>= lim_{T\rightarrow∞}<A(T)|B(-T)> = lim_{T\rightarrow∞}<A|e^{-iH(2T)}|B> [/itex]
    In the last the states are taken at the common reference time.
    In that case you get a relation between incoming and outcoming states- related by the limit of a sequence of unitary operators. The limiting unitary operator is the S-matrix.
    [itex] <A_{out}|B_{in}>= <A|S|B> [/itex]
     
  8. May 11, 2014 #7
    Okay, I think I see the difference. It's confusing, and maybe I still have it wrong, but this is what I have.

    First, I think it's most natural to view the S-matrix as an operator that maps an in-state to some linear combination of out-states:

    $$\langle \beta_{out} | S|\alpha_{in} \rangle=\langle \beta_{out} | S_{\gamma \alpha} \gamma_{out} \rangle =S_{\beta \alpha}$$

    (repeated indices are summed)
    But other textbooks say that the in-state "is" a linear combination of out-states. This would look like:

    $$\langle \beta_{out} |\alpha_{in} \rangle =\langle \beta_{out} | S_{\gamma \alpha} \gamma_{out} \rangle=S_{\beta \alpha} $$

    They give the same result, either viewpoint. The point is that in the first expression, the indices on S are two different indices. The 2nd index represents the in-state, and the 1st represents the out-state. So the operator S in this case has two different basis. I don't see why this would be a problem, but usually in QM an operator is expressed in a single basis with both indices. For example you don't consider elements of the operator O like <p|O|x>, or <x|O|p>, but rather the same basis: <q|O|p> or <x_2|O|x_1>, i.e., all momentum or all position, and no mixed. But still it seems legitimate to have things mixed: O= |p><p|O|x><x|

    With the 2nd viewpoint, you don't have an operator, but a linear combination which represents the fact that the in-state "is" a linear combination of the out-state. But if we insist on having an operator that has the same basis for both indices, we can write:

    $$\langle \beta_{out} |\alpha_{in} \rangle=\langle \beta_{out} | S_{\gamma \alpha} \gamma_{out} \rangle=
    \langle \beta_{out} | \gamma_{out} \rangle S_{\gamma \alpha}=
    \langle \beta_{out} | \gamma_{out} \rangle S_{\gamma \eta} \langle \eta_{out} | \alpha_{out}\rangle=\langle \beta_{out} | \big[ |\gamma_{out} \rangle S_{\gamma \eta} \langle \eta_{out}| \big]| \alpha_{out}\rangle$$

    The operator in [] is the S-matrix, but with both indices now out-states.

    Numerically the S-operators are equal in their respective basis, but one of the basis is mixed. The 1st S-operator takes an in-state and give you a linear combination of out-states, and the 2nd S-operator takes an out-state and gives you a linear combination of out-states (which is equal to the in-state!).

    Anyways, I think I'm still confused lol.
     
  9. May 11, 2014 #8
    As you said, this is a rather complicated topic. However I think you are making a little bit of confusion between the action of the S matrix on the in state (first case) and its definition (second case). The S matrix is defined as the matrix that tells you what mixture of out states your in state must be.

    There is no doubt that your initial in state must be a combination of out states. In facts, when you use the formalism with in and out states, you don't use any evolution operator: the in state already contains all the possible final states it can go into.

    This is not true. In QM you take scalar products between different basis all the time. The most common example is the free particle wave function. It is exactly the scalar product between the ket in the momentum basis and the bra in the position basis:
    $$
    \langle x|p\rangle=\frac{e^{ipx/\hbar}}{\sqrt{2\pi\hbar}}.
    $$
     
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