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Expectation values for an harmonic oscillator

  1. Jun 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the expectation values of x and p for the state
    [itex]\vert \alpha \rangle = e^{-\frac{1}{2}\vert\alpha\vert^2}exp(\alpha a^{\dagger})\vert 0 \rangle[/itex], where ##a## is the destruction operator.

    2. Relevant equations
    Destruction and creation operators
    ##a=Ax+Bp##
    ##a^{\dagger}=Ax-Bp##
    For some constants A (real) and B (imaginary) whose value is not important now.
    3. The attempt at a solution
    I've found a solution, but it is so simple it looks dumb. State the expectation value:
    ##\langle x \rangle = \langle 0 \vert e^{- \frac{1}{2}\vert\alpha \vert ^2} e^{\alpha^*a}\hat{x}e^{\alpha a^{\dagger}}e^{- \frac{1}{2}\vert\alpha \vert ^2} \vert 0\rangle = ##

    Now if we descompose the first exponential ##e^{\alpha^*a}## there will be an annihilation operator acting on the ground state, which is equal to 0 and therefore ANY expectation value for this state is zero...

    That seems wrong. Where did I go wrong?
     
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  3. Jun 28, 2014 #2

    Orodruin

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    ##a## is never acting directly on a ground state ket. How do the creation and annihilation operators act on bra states?
     
  4. Jun 28, 2014 #3
    I was not sure of this, but I still don't see the mistake. Look at what the expression for the expectation value I wrote. There is an exponential of the annihilation operator next to the ground state; if we decompose it... oh, wait.
    Okay, I just realized that the first term of the series would not contain an ## a ## operator, so is doesn't destroy it.

    New try:

    From the definition for ##a## I've found ##x=\frac{a+a^{\dagger}}{2A}## and ##p=\frac{a-a^{\dagger}}{2B}##, and then its easy to find:

    ##\langle x \rangle = \langle 0 \vert e^{- \frac{1}{2}\vert\alpha \vert ^2} e^{\alpha^*a}\hat{x}e^{\alpha a^{\dagger}}e^{- \frac{1}{2}\vert\alpha \vert ^2} \vert 0\rangle = \frac{1}{2A} \langle 0 \vert e^{ -\frac{1}{2}\vert\alpha \vert ^2} e^{\alpha^*a}(a + a^{\dagger})e^{\alpha a^{\dagger}}e^{-\frac{1}{2}\vert \alpha \vert ^2} \vert 0\rangle ##

    Now this reduces to ##\langle x \rangle = \frac{1}{2A} \left ( \langle a \rangle + \langle a^{\dagger} \rangle \right )## and, operating similarly for the momentum operator ##\langle p \rangle = \frac{1}{2B} \left ( \langle a \rangle - \langle a^{\dagger} \rangle \right )##.

    Now, I think ##\langle a \rangle = \langle a^{\dagger} \rangle ##, is that right?
     
  5. Jun 28, 2014 #4

    Orodruin

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    Neither operator in ##\hat x## act directly on the vacuum. They must first be commuted through the exponential expressions that contain operators.

    By definition, we must have
    $$
    \langle a\rangle = \langle a^\dagger\rangle^*
    $$
     
  6. Jun 28, 2014 #5
    Yes it seems right at first look. So ##<p>=0##. What about ##<a>##?

    [edit: oops yes ##<>^*##]
     
    Last edited: Jun 28, 2014
  7. Jun 28, 2014 #6
    I know from a previous exercise that these states are eigenkets of ##a## with eigenvalue ##\alpha##, so ##\langle a \rangle = \alpha ## and ##\langle x \rangle = \frac{\alpha}{A}##?
     
  8. Jun 28, 2014 #7

    Orodruin

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    Is ##\alpha## necessarily real?
     
  9. Jun 28, 2014 #8
    No. Why?
     
  10. Jun 28, 2014 #9

    Orodruin

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    If ##\alpha## is not real, is p zero?
     
  11. Jun 28, 2014 #10
    Oh, thank you. The actual solutions are
    ##\langle \hat{x} \rangle = \frac{Re(\alpha)}{A}##
    ##\langle \hat{p} \rangle = \frac{Im(\alpha)}{B}##
    The momentum EV is actually real, because B is pure imaginary.
     
  12. Jun 29, 2014 #11
    Hi Carllacan - not sure if you figured this out or not - but anyway here's the cleanest way I can think of to do it:

    [tex]\begin{eqnarray}
    \langle X \rangle &=& e^{-\vert \alpha \vert^2} \frac{1}{2A} \langle 0 \lvert e^{\alpha^*a}(a + a^{\dagger}) e^{\alpha a^{\dagger}} \lvert 0\rangle\\

    &=& \frac{1}{2A} e^{-\vert \alpha \vert^2} \left( \frac{\partial}{\partial \alpha} + \frac{\partial}{\partial \alpha^*}\right) \langle 0 \lvert e^{\alpha^*a} e^{\alpha a^{\dagger}} \lvert 0\rangle\\

    &=& \frac{1}{2A} e^{-\vert \alpha \vert^2} \left( \frac{\partial}{\partial \alpha} + \frac{\partial}{\partial \alpha^*}\right) e^{\alpha^* \alpha}\\

    &=& \frac{1}{2A} e^{-\vert \alpha \vert^2} \left( \alpha + \alpha^* \right) e^{\alpha^* \alpha}\\

    &=& \frac{\alpha + \alpha^*}{2A}\\

    &=& \frac{{\rm Re} \; \alpha}{A}

    \end{eqnarray}

    [/tex]
    I'll let you do [itex]\langle P \rangle[/itex] for yourself :smile:

    Here's a couple of "for what it's worth" comments:

    The state [itex]\lvert \alpha \rangle[/itex] is called a "canonical coherent state" and uses the harmonic oscillator ground state as what's called its "fiducial vector". However note that there need not be a harmonic oscillator present - our particle could be in absolutely any potential.

    The state [itex]\lvert \alpha \rangle[/itex] has (the expectation values of) its position and momentum coded into the real and imaginary parts of [itex]\alpha[/itex]. So we could rewrite it as simply [itex]\lvert p,q \rangle[/itex]. Thus there's a nice one-to-one map from classical phase space onto canonical coherent states in the Hilbert space - it's as good as a map as the uncertainty principle allows.
     
  13. Jun 29, 2014 #12
    Thank you, I was wondering what was the point of having us work with these specific states.

    Would you guys mind checking this?
    We know ##a\vert \alpha\rangle = \alpha \vert \alpha \rangle ##, so
    ##\langle a \rangle = \alpha##,
    ##\langle a^{\dagger} \rangle = \langle a \rangle ^* = \alpha^*##,
    ##\langle a^2 \rangle = \langle \alpha \vert a^2 \vert \alpha \rangle = \langle \alpha \vert \alpha^*\alpha \vert \alpha \rangle = \alpha ^* \alpha = \vert \alpha\vert ^2##,
    ##\langle a_{\dagger}^2 \rangle = \langle a^2 \rangle ^* = \vert \alpha\vert ^2##,
    ##\langle a a_{\dagger} \rangle = \langle \alpha \vert a a_{\dagger} \vert\alpha \rangle = \alpha \langle \alpha \vert a_{\dagger} \vert\alpha \rangle = \vert \alpha\vert ^2##
     
  14. Jun 29, 2014 #13
    Well the notation [itex]\langle a\rangle[/itex] looks a bit weird to me, because it suggests you're finding the expectation value of some observable - but [itex]a[/itex] isn't an observable. I guess it's ok though.

    I would say that:

    (I'm changing [itex]\alpha[/itex] to [itex]\lambda[/itex] because with my eyesight I can't tell the difference between an [itex]\alpha[/itex] and an [itex]a[/itex] on the screen :frown:)

    [tex]\begin{eqnarray}
    \langle \lambda \lvert a \lvert \lambda \rangle &=& \lambda\\
    \langle \lambda \lvert a^{\dagger} \lvert \lambda \rangle &=& \lambda^*\\
    \langle \lambda \lvert a^2 \lvert \lambda \rangle &=& \lambda^2\\
    \langle \lambda \lvert (a^{\dagger})^2 \lvert \lambda \rangle &=& (\lambda^*)^2\\
    \langle \lambda \lvert a^{\dagger}a \lvert \lambda \rangle &=& \lambda^*\lambda\\
    \end{eqnarray}[/tex]
    So I disagree with you about some of them at least.
     
  15. Jun 29, 2014 #14

    Orodruin

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    I agree with Oxvillian's results. I would however not write the coherent state as ##|p,q\rangle## without some additional hint to it being a coherent state. Otherwise there is a risk of interpreting it as an eigenstate of the position and momentum operators (there are no simultaneous eigenstates, but anyway...)
     
  16. Jun 29, 2014 #15
    Indeed, [itex]\lvert p,q \rangle[/itex] looks like a total violation of page 1 of the quantum mechanics rulebook :smile:

    You just have to remember that the labels [itex]p[/itex] and [itex]q[/itex] are just parameters, not eigenvalues.

    Anyway when you get past the notation, there's a lot to be said for working in the [itex]\lvert p,q \rangle[/itex] basis instead of using [itex]\vert p \rangle[/itex] and/or [itex]\vert q \rangle[/itex]. The coherent states are nice and smooth rather than delta-function spikey. They lead, for instance, to much nicer path integrals.
     
  17. Jun 29, 2014 #16
    About the third one: doesn't ##a\vert \lambda \rangle = \lambda \vert \lambda \rangle ## imply ## \langle \lambda\vert a^{\dagger} = \langle \lambda \vert \lambda^* ##?

    And therefore the result is ##\lambda ^* \lambda = \vert \lambda \vert ^2##
     
  18. Jun 29, 2014 #17
    That would be the 5th one.

    The 3rd one is
    [tex]\langle \lambda \lvert aa \lvert \lambda \rangle = \lambda \langle \lambda \lvert a \lvert \lambda \rangle = \lambda^2 \langle \lambda \lvert \lambda \rangle = \lambda^2[/tex]
     
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