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Klien-Gordon equation to solve 4-vector problem? (Particle)

  1. Feb 22, 2015 #1
    1. The problem statement, all variables and given/known data
    kU1D3nQ.jpg

    2. Relevant equations
    KG Equation

    ##\left ( \delta ^{2} + \frac{m^{2}c^{2}}{\hbar^{2}}\right )\Psi = 0##

    3. The attempt at a solution

    To solve this problem I'm not sure whether to use the KG equation OR apply the operator given in the question (I dont know how to do this)

    I can use the KG equation on PSI when given say:

    ##\Psi = R exp(-iwt+ik_{i}x_{i})##

    But if I use the KG equation for this problem, so treat Aμ as PSI in the above equation, I will need to write it in non-covarient form. To do this I can simply take the inner product of qv and xv but I dont know what they are equal to, to be able to do this.

    I'm a bit lost!

    Please could someone suggest a starting point or at least give me an idea of what qv and xv are actually equal to so I can attempt to apply the operator (any hints on how to do this would be appreciated) or the KG equation.

    Thanks for any suggestions in advance.

    This is a past exam question that I'm attempting to apply my lecture notes to, it's not any assessed piece of work.
     
  2. jcsd
  3. Feb 22, 2015 #2

    TSny

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    The components ##q_{\nu}## are just constants. Suppose you apply ##\hat{p}_1## to ##A^\mu##. What do you get?
     
  4. Feb 22, 2015 #3
    thats the problem I dont know what these things are, I know

    ##\delta _{\mu}= (\frac{1}{c}\frac{\delta}{\delta t},\frac{\delta}{\delta x},\frac{\delta}{\delta y},\frac{\delta}{\delta z})##

    and

    ##\rho_{\mu}= (\frac{E}{c},-\rho_{x},-\rho_{y},-\rho_{z})##

    and

    ##A^{\mu}= (\frac{\Phi }{c},A_{x},A_{y},A_{z})##

    but what would δk even be?

    i'm struggling with what each of the things in the question are actually equal to.

    thanks for the help
     
    Last edited: Feb 22, 2015
  5. Feb 22, 2015 #4

    TSny

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    You want to show that ##A^{\mu}## is an eigenfunction of each of the operators ##\hat{p}_\kappa##. So, take the case where ##\kappa = 1##. Then ##\hat{p}_1 =i \hbar \frac{\partial}{\partial x}##. Let this act on ##A^{\mu}##. It will help if you explicitly write out ##q_{\nu}x^{\nu}## in the exponential factor of ##A^{\mu}##.
     
    Last edited: Feb 23, 2015
  6. Feb 23, 2015 #5
    thanks again for the reply. but that's where my problem lies, I don't know what ##q_{\nu}x^{\nu}## explicitly are. what do ##q_{\nu}## and ##x^{\nu}## equal individually? sorry I'm very new to the notation and we have yet to be given any problems in this notation.
     
  7. Feb 23, 2015 #6

    TSny

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    The notation ##q_{\nu}x^{\nu}## means ##\sum_{\nu = 0}^3q_{\nu}x^{\nu} = q_0x^0 +q_1x^1+q_2x^2+q_3x^3##. The summation is implied in the repeated indices of the index ##\nu## in ##q_{\nu}x^{\nu}##. This is the "Einstein summation convention".

    The various ##q##'s are constants whose meaning you will discover in this exercise. ##R## and ##\epsilon^{\mu}## are also constants.
     
  8. Feb 24, 2015 #7
    thanks very much, I will try with this info and see how far I get.
     
  9. Feb 26, 2015 #8
    ok, this is easier than i thought (i think), I'll post where I am incase it helps others.

    ##\delta ^{\mu} = \frac{\delta }{\delta x^{\mu}}##

    and

    ##\rho ^{v}x_{v} = \rho^{0}x_{0}+\rho^{1}x_{1}+ \rho^{2}x_{2} + ...##

    you can see from this that x is linear, therefore for any term of xv the derivative with respect to x will be 1.

    therefore

    ##\delta ^{\mu} \rho ^{v}x_{v} = \frac{\delta }{\delta x^{\mu}} \rho ^{v}x_{v} = \rho^{v} \frac{\delta }{\delta x^{\mu}} x_{v}=\rho^{v}##

    edit

    hmm so going back to the problem, it would be:

    ##p_{k}A^{\mu} = i \hbar(R\epsilon ^{\mu})\frac{\delta }{\delta x^{k}}(e^{-iq_{v}x^{v}}) ##

    in this case would

    ##\frac{\delta }{\delta x^{k}} x^{v} = 1##?

    as the v is an indices and doesnt mean x to the power right?

    any further advice would be appreciated
     
    Last edited: Feb 26, 2015
  10. Feb 26, 2015 #9

    TSny

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    This is not quite right. Note that the index ##\nu## is a summation index, so the result cannot be ##\rho^{v}##. It would be best to raise and lower the ##\nu## index to write it as ## \frac{\delta }{\delta x^{\mu}} \rho _{v}x^{v}##. Try to show that ##\frac{\delta }{\delta x^{\mu}} x^{v} = \delta_{\mu \nu}## where ##\delta_{\mu \nu}## is the Kronecker delta. Or, if you have trouble with this, write out the sum over ##\nu## explicitly:

    ## \frac{\delta }{\delta x^{\mu}} \rho _{v}x^{v} = \frac{\delta }{\delta x^{\mu}} ( \rho _{0}x^{0}+ \rho _{1}x^{1} + \rho _{2}x^{2} + \rho _{3}x^{3})##


    It might help to write out the argument of the exponential: ##p_{k}A^{\mu} = i \hbar(R\epsilon ^{\mu})\frac{\delta }{\delta x^{k}}(e^{-i( q_0x^0 + q_1x^1 + q_2x^2 + q_3x^3 }) ##
     
  11. Feb 27, 2015 #10
    damn, though I had it. ok I'll ask a last question and go from there as you have already helped plenty.

    ##\frac{\delta }{\delta x^{\mu}} x^{v} = \delta_{\mu \nu}##

    this would mean that when mu=v the term would be 1 and zero for non equal componants. so

    ##\frac{\delta }{\delta x^{1}}x^{2} = 0##

    im now getting confused with indices and powers, could you confirm that x2 doesnt mean x squared, it's just a label for the 2nd component of xv?

    so you could write it ##\frac{\delta }{\delta x^{1}}x^{1} = \frac{\delta }{\delta x^{1}}(x^{1})= \frac{\delta }{\delta A}(A)=1##

    or likewise ##\frac{\delta }{\delta x^{2}}x^{2} = \frac{\delta }{\delta x^{2}}(x^{2})= \frac{\delta }{\delta B}(B)=1##
     
  12. Feb 27, 2015 #11

    TSny

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    OK. So, for example, ##\frac{\partial}{\partial x^k}(q_0x^0+q_1x^1+q_2x^2+q_3x^3) = q_k## for k = 0, 1, 2, or 3.
     
  13. Feb 28, 2015 #12
    Thanks for all your help. When I finally get the question finished I'll be sure to post here.
     
  14. Mar 1, 2015 #13
    sorry please delete
     
    Last edited: Mar 1, 2015
  15. Mar 1, 2015 #14
    Ok think I'm there, I'll write it in detail (including the small things I've picked up) edit i've written Au wrong! but I believe the method is correct for the Au i've written, the original question would require a switch of indices

    ##A^{\mu} = R\epsilon^{\mu} e^{-k^{v}x_{v}}##

    ##p_{k}=i\hbar\delta _{k}=i\hbar \frac{\delta }{\delta x_{k}}##

    note i originally wrote this ##p_{k}=i\hbar\delta _{k}=i\hbar \frac{\delta }{\delta x^{k}}## which would have been ##p^{k}## not ##p_{k}##

    therefore since the k indices of the differential matches that of the indices of the x componant in the exponential there is no need to switch indices.

    ##p_{k}A^{\mu} =i\hbar R\epsilon^{\mu}\frac{\delta }{\delta x_{k}} e^{-k^{v}x_{v}}##

    ##=i\hbar R\epsilon^{\mu} e^{-k^{v}x_{v}}\frac{\delta }{\delta x_{k}} (-k^{v}x_{v})##

    ##=i\hbar R\epsilon^{\mu} e^{-k^{v}x_{v}} (-ik^{k})\frac{\delta }{\delta x_{k}} (-k^{v}x_{v})##

    note that when the differential acts on the term that it changes it's indices to match the indices letter of the differential.

    ##=i\hbar (-ik^{k}) A^{\mu}=\hbar k^{k} A^{\mu}##
     
    Last edited: Mar 1, 2015
  16. Mar 1, 2015 #15

    TSny

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    Your original way of writing ##\hat{p}_k## was correct. That is ##\hat{p}_k = i\hbar\partial_k = i\hbar\frac{\partial}{\partial x^k}##.

    Note that the ##x^k## is in the denominator. So, the superscript ##k## on ##x^k## actually acts like a subscript for the overall expression. That is why the ##k## is a subscript on ##\hat{p}## while it is a superscript on ##x##.

    I don't understand the sudden appearance of ##(-ik^{k})## in the last line.
     
  17. Mar 1, 2015 #16
    hmm I didnt know that, thanks. So what would ##\hat{p}^{k}## be? ##\hat{p}^{k} = i\hbar\partial^{k} = i\hbar\frac{\partial}{\partial x_k}## which would then act on the upper indices?

    this is what was given in class:

    IrsJu9a.jpg

    it doesnt show the steps but you can see that after acting on the exponential the differential brings down a ##k^{\rho}## not a ##k^{v}##. So i've applied the same idea to the ##k^{v}## to give ##k^{k}##
     
  18. Mar 1, 2015 #17

    TSny

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    Yes.

    ##\frac{\partial}{\partial x_k}## would act on any function of ##x_k## (with lower index on ##x##).

    That looks good.

    You had
    You left out the ##i## in the arguments of the exponentials. You are correct that ##\frac{\delta }{\delta x_{k}} (-ik^{v}x_{v}) = -ik^k##

    In your last line you should not still have the factor ##\frac{\delta }{\delta x_{k}} (-k^{v}x_{v})##
     
  19. Mar 2, 2015 #18
    oops, indeed.

    I am very happy to leave it there, thanks for teaching me 4-vectors!

    There is just one final thing, this statement:

    if ##\frac{\partial}{\partial x_k}## acts on the lower indices and ##\frac{\partial}{\partial x^k}## acts on the lower indices, then what's the difference between them? I don't understand that they both act on the lower indices of x.
     
  20. Mar 2, 2015 #19

    TSny

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    ##\frac{\partial}{\partial x_k}## and ##\frac{\partial}{\partial x^k}## do not act on indices, they act on functions of ##x_k## or ##x^k##.

    I'm not sure which convention of the metric you are using. But let's suppose you are using the metric where ##x_0 = x^0## and ##x_i = - x^i## for i = 1, 2, or 3. Then, for example, $$\frac{\partial x^1}{\partial x^1} = 1$$ but $$\frac{\partial x^1}{\partial x_1} =\frac{\partial (-x_1)}{\partial x_1} = - \frac{\partial x_1}{\partial x_1} =-1$$

    As a little exercise, you can try evaluating $$\frac{\partial }{\partial x^3}(x_{\mu}x^\mu) $$ and $$\frac{\partial }{\partial x_3}(x_{\mu}x^\mu) $$
     
  21. Mar 2, 2015 #20

    Will try that. Thanks once again for all your help!
     
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