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Quantum: why can't you square every operator?

  1. Sep 29, 2013 #1
    In Griffiths book, he says (a+ + a-)2 = a+2 + a-2 + a+a- + a-a+.

    Why can you NOT do the same thing for a+2 = (-ip+mωx)2 ?!

    When I do this to find the 2nd excited state of SHO, it gives me wrong answer. I actually have to apply a+ two times to ψ0 in order to get ψ2. It is ridiculous that Griffiths does not mention at all why certain operators can be squared but others cannot. Could someone please elaborate on the difference between these two cases?
  2. jcsd
  3. Sep 29, 2013 #2


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    I hope you realize that squaring an operator means applying the operator twice.
  4. Sep 29, 2013 #3
    In the case of
    (a+ + a-)2 = a+2 + a-2 + a+a- + a-a+.

    you are not applying it twice. You are squaring the operator and then applying that result on ψ. That is exactly what griffiths did.
  5. Sep 29, 2013 #4


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    By definition, ##A^2 = A \circ A## for some operator ##A## assuming that the domains agree (which is only a problem if ##A## is unbounded).
  6. Sep 29, 2013 #5


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    Let me elaborate: let ##\psi## be an arbitrary vector in the domain of ##(a_+ + a_-)##-we take for granted that the action of ##(a_+ + a_-)^2## is well defined on ##\psi##. Then, ##(a_+ + a_-)^2 \psi = (a_{+} + a_{-})[(a_{+} + a_{-})\psi] = (a_+ + a_-)[a_+ \psi + a_- \psi] = a_+^2 \psi + a_- a_+ \psi + a_+a_- \psi + a_-^2 \psi##.
  7. Sep 29, 2013 #6

    you are not answering my question. sure, i can agree that in general, A^2=A*A. But that is not my question. My confusion is that for certain cases (idk if it depends on the operators involved or the wavefunction involved), you can directly FOIL A^2, as in the above example i already mentioned twice. My question is therefore, why can you FOIL certain operators, but not others.

    If this is still not clear, let me rephrase again: In the above post, you have two operators:

    A = (a+b)^2 = (a+ + a-)^2
    B = (c+d)^2 = (-ip + mωx)^2

    For the operator A, you can FOIL it, and apply the result on ψ.
    For the operator B, you CANNOT FOIL it, and apply the result on ψ.

    Thus my question is, what is the difference between A and B that causes this asymmetry?!
  8. Sep 29, 2013 #7


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    What did you get when you tried foiling ##(-iP + m\omega X)^{2}## and acting on some vector?
  9. Sep 29, 2013 #8
    Idk why we would take anything for granted, but either way, this is still not how griffiths did this. You are applying the operators twice on ψ. griffiths did not do this. He literally said that A^2 = (FOIL'ed A^2). THEN he applied this result on ψ. Thus from this example, I thought the message was that you can just FOIL operators.

    Are you saying then that this procedure is always wrong? Or are there still certain operators that can be exponentiated before being applied?
  10. Sep 29, 2013 #9
    I got the wrong answer when acting on the vector.

    I FOIL'ed it as:

    -h_bar^2 * (d^2/dx^2) + m^2ω^2x^2 - m*h_bar*ω*x*(d/dx) - m*h_bar*ω
  11. Sep 29, 2013 #10
    I think the effective Socratic question here is to ask you what you think an operator is. I know relatively little about quantum physics, but math will always whisper answers in one's ear if you are willing to listen.
  12. Sep 29, 2013 #11
    The last term you forgot that the product rule will be invoked. (d/dx) x is not just 1 because d/dx will also act on the state.
  13. Sep 29, 2013 #12


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    It's a mathematical technicality. If ##A## is an unbounded operator then it's domain is not going to be the entire vector space so ##A\psi## need not be a vector in the domain of ##A## meaning ##A^2 \psi## will not be well defined if that happens.

    You didn't foil correctly. You have to be extra careful when foiling because you're dealing with operators here. It should be ##(-iP + m\omega X)^2\psi(x) \\= (-iP)^2 \psi(x) + (m \omega X )^2\psi(x) - (im\omega PX)\psi(x) - (im\omega XP)\psi(x) \\= \hbar^2 \frac{\mathrm{d} ^2 \psi}{\mathrm{d} x^2} + m^2\omega^2 x^2 \psi - (m\omega \hbar \frac{\mathrm{d} }{\mathrm{d} x})(x\psi(x)) - m\omega \hbar x\frac{\mathrm{d} \psi}{\mathrm{d} x} \\= \hbar^2 \frac{\mathrm{d} ^2 \psi}{\mathrm{d} x^2} + m^2\omega^2 x^2 \psi - 2m\omega \hbar x\frac{\mathrm{d} \psi}{\mathrm{d} x} - m\omega \hbar \psi##
  14. Sep 29, 2013 #13
    oohhhh! yes you are correct, thank you so much! so then i CAN exponentiate any operator, but i have to remember to immediately multiply by my function.

    thank you!!
  15. Sep 29, 2013 #14
    For clarity: Griffiths suggests that you do this whenever you compute a commutator. You can extend that rule of thumb to any operator computation: always use a test function for operator equations.
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