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Propagator equation

  1. Jul 14, 2007 #1
    Can someone explain to me why the equation [tex] \psi(x,t) = \int{U(x,t,x',t')\psi(x',t')dx'} [/tex] where U is the propagator has an integral? I thought you could just multiply the propogator by an initial state and get a final state?
  2. jcsd
  3. Jul 14, 2007 #2
    In quantum mechanics, time evolution of wave functions is represented by the time evolution operator. The formula you wrote is the most general linear operator connecting wave functions at times t and t'.
  4. Jul 14, 2007 #3
    Perhaps so, if you call the time evolution operator a propagator. You can get the time evolution by multiplying the state vector by a correct operator

    |\psi(t)\rangle = e^{-iH(t-t')/\hbar}|\psi(t')\rangle

    This is quite abstract like this. If you use the position representation, then this operator is something more complicated that just a function, and this "multiplication" is not a pointwise multiplication of two functions.
  5. Jul 14, 2007 #4
    OK. So for the position representation (which is the same as the X basis, right?) we have that (for a free particle):

    [tex]U(x,t;x') \equiv <x|U(t)|x'> = \int^{\infty}_{\infty}<x|p><p|x'>e^{-ip^2/2m\hbar}dp [/tex]

    which can be reduced to [tex](\frac{m}{2\pi\hbar i t})^{1/2}e^{im(x-x')^2/2\hbar t} [/tex].

    So why is it not [tex]\psi(x,t) = (\frac{m}{2\pi\hbar i t})^{1/2}e^{im(x-x')^2/2\hbar t} \psi(x',0)[/tex]

    or simply [tex]\psi(x,t) = U(x,t;x') \psi(x',0)[/tex]?

    Also, what exactly does the equivalence mean [tex]U(x,t;x') \equiv <x|U(t)|x'> [/tex]?
    Last edited: Jul 14, 2007
  6. Jul 14, 2007 #5
    I could well be wrong, and at best I'll give a very restricted view since I probably know less than you, but...

    Think about a particle with psi(x,t=0) a delta function. Over time psi spreads out, and psi(x',t)=U(x,t;x') psi(x,t=0).
    But then think about (the maybe unphysical I don't know) situation of a particle starting with two inifinitly thin peaks. Then at a later time, at position x', psi will be given by the contribution from the first peak which has spread out + the contribution from the second peak spread. Generalise for a continuous wavefunction and you get an integral.
  7. Jul 14, 2007 #6
    That makes sense except I think you may have mixed up your primes in the expression for U(t) but still that explanation of the integral really helped.
  8. Jul 14, 2007 #7
    This is an amplitude for the particle to move from point x' to point x in time t.

    Because the particle can arrive to the point x not only from x', but from any other point in space as well. So, this expression should be integrated on x' in order to get the full amplitude of finding the particle at point x.

    I think that explanation given by plmokn2 is a good one.

  9. Jul 15, 2007 #8
    So I see how you can think of its as U(t) operating on the ket |x'> to get U(t;x') but how can you think of the relationship between the bra <x'| and the operator U(t;x'). What does <x'|U(t;x') "mean"?
  10. Jul 15, 2007 #9
    In the notation U(t;x, x') = < x| U(t) |x'> you can first apply the unitary operator U(t) to the ket vector |x'> on the right and obtain a new ket vector
    (which is a result of a time translation applied to |x'>), which I denote by

    |x', t> = U(t) |x'>

    In the next step you can take an inner product of |x', t> with the bra vector <x|

    U(t;x, x') = <x |x', t>

    U(t;x, x') is a complex number which can be interpreted as a matrix element of the unitary operator U(t) in the basis provided by vectors |x>.
  11. Jul 15, 2007 #10
    ehrenfest, you should do the exercise, where Shrodinger's equation is derived out of a time evolution defined with a propagator. After it, it becomes easier to believe in propagators, and in how they work. If the sources you are using don't explain it, you can get hints from here.
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