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Particle in coherent state

  1. Dec 2, 2014 #1
    1. The problem statement, all variables and given/known data
    A particle in harmonic potential $$H=\hbar \omega (a^\dagger a+1/2)$$ is at ##t=0## in coherent state $$a|\psi>=2|\psi>$$ a) Calculate expected value of energy and it's uncertainty at ##t=0##
    b) With what probability do we at ##t=0## measure the energy less than ##3\hbar \omega##?
    c) And what if we measure it at ##t=\frac \pi \omega##?

    2. Relevant equations


    3. The attempt at a solution
    I have to apologize, but I only have an idea on have to try with the a) part, while I haven't got a clue on b) and c).
    a)$$<H>=<\psi |H|\psi>=\hbar \omega( <\psi|\psi>+<\psi|a^\dagger a|\psi>$$ I hope I can say that ##<\psi|a^\dagger a|\psi>=2<\psi|a^\dagger |\psi>=2<a\psi|\psi>=4<\psi|\psi>=4##

    Than ##<H>=\frac 9 2 \hbar \omega##. $$<H^2>=(\hbar \omega)^2(<\psi|(a^\dagger a)(a^\dagger a)|\psi>+<\psi|a^\dagger a|\psi>+\frac 1 4)$$
    Now I am not sure what to do with ##<\psi|(a^\dagger a)(a^\dagger a)|\psi>##... :/

    b) and c).. I would be really happy to get a hint here :/
     
  2. jcsd
  3. Dec 2, 2014 #2

    Orodruin

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    (a) What kind of relations do you know that lets all of the annihilation operators act to the right and all the creation operators to act to the left?

    (b) What states have energy less than this? What is the probability of being in one of these states?

    (c) Same as (b) with the additional question: How does the state evolve with time?
     
  4. Dec 3, 2014 #3
    a) I didn't quite understand your question, but I did manage to get a solution, hopefully without mistakes: $$<H>=\frac 9 2 \hbar \omega $$ and $$<H^2>=<\psi|(a^\dagger a+1/2)^2|\psi>=(\hbar \omega)^2(<\psi|a^\dagger (aa^\dagger) a|\psi>+<\psi |a^\dagger a|\psi>+\frac 1 4<\psi|\psi>)$$ It is useful to realize that ##[a,a^\dagger]=1## because this tells me that ##aa^\dagger =1+a^\dagger a##. Using this gives me
    ##<\psi |a ^ \dagger (aa ^ \dagger ) a | \psi>## ##=<\psi |a ^ \dagger a| \psi >+## ##<\psi |a ^ {\dagger 2} a^2| \psi >=##
    ##=2<\psi | a ^ \dagger \psi >+## ##4<\psi |a ^{\dagger 2}| \psi >## ##=2<a \psi | \psi >+## ##4<a ^ {\dagger 2} \psi | \psi >## ##=4+16##
    And the second and third term in Hamiltonian together are $$<\psi |a^\dagger a|\psi>+\frac 1 4<\psi|\psi>=4+\frac 1 4 $$ Therefore $$<H^2>=(\hbar \omega) ^2 \frac{97}{4}$$.

    Would this be ok?

    b) Yes I thought about this. Generally speaking, what Hamiltonian does is $$H|\psi>=\hbar \omega(n+1/2)|\psi>$$ which leaves me with 3 possible states ##n\in {0,1,2}## if ##E\leq 3\hbar \omega##. BUT probability of being in any of those states is what I am having troubles with, because If I am not mistaken all states are possible.
     
    Last edited: Dec 3, 2014
  5. Dec 3, 2014 #4

    Orodruin

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    Yes, this is what I was referring to.

    (b) What you write is true for the energy eigenstates, not for the general state ##|\psi\rangle##. What is the probability of being in any eigenstate if you know the wave function of the current state and the eigenfunctions of the Hamiltonian?
     
  6. Dec 3, 2014 #5
    b) Hmmm... Well, a general state ##|\psi>## could be (or is?) a superposition of eigenstates ##|\psi>=\sum _n C_n|n>##, if so, than ##|C_n|^2## is the probability to be in ##n## eigenstate.
    And I am assuming this is not what you wanted to hear, because this doesn't help at all. :D
     
  7. Dec 3, 2014 #6

    Orodruin

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    It is, the question is how you are going to extract the ##C_n## coefficients.
     
  8. Dec 3, 2014 #7
    And how can I do that?
     
  9. Dec 3, 2014 #8

    Orodruin

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    If you had a vector you can write it as a linear combination of the basis vectors, how would you extract the ith component of the vector? (I.e., the coefficient of the ith basis vector)
     
  10. Dec 3, 2014 #9
    Multiply everything with the ith basis vector.
     
  11. Dec 3, 2014 #10

    Orodruin

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    Exactly. Therefore ...
     
  12. Dec 3, 2014 #11
    ##|\psi>=\sum _n C_n|n> \Rightarrow <n|\psi>=C_n##

    But I would need some numbers, right? Why do I have a feeling that this is supposed to be really simple and I just can't see it?
     
  13. Dec 3, 2014 #12

    Orodruin

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    How do you express the inner product between two states in position basis?
     
  14. Dec 3, 2014 #13
    ##<n|m>=\int \psi_n\psi_m^*=\delta _{n,m}##

    Position basis? I think I have never done that. Does this add ##cos(\vartheta )## in the equation above?
     
  15. Dec 3, 2014 #14

    Orodruin

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    You probably have but you might not have called it that, it is the most common basis where you represent the states by their wave functions.
     
  16. Dec 4, 2014 #15
    Ok maybe, just maybe I got it now.

    We can expand coherent states in Fock space $$|z>=e^{-\frac 1 2 |z|^2}\sum_{n=0}^\infty \frac{z^n}{\sqrt{n!}}|n>$$ which in my case for ##n\in
    \begin{Bmatrix}
    0,1,2
    \end{Bmatrix}## means $$e^{-2}|0>+2e^{-2}|1>+\frac{4}{\sqrt 2}e^{-2}|2>$$ This than gives me the desired coefficients and possible energies.
     
  17. Dec 4, 2014 #16

    Orodruin

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    Correct, but note that if you are not given this form of a coherent state, you must first show that you can write the expansion on this form.
     
  18. Dec 4, 2014 #17
    THANK YOU! (also for some other problems I had)
     
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