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I Simple calc. of energy levels in quantum harmonic oscillator

  1. Aug 2, 2016 #1
    Derivation of energy levels in a quantum harmonic oscillator, ##E=(n+1/2) \hbar\omega##, is long, but the result is very short. At least in comparision with infinite quantum box, this result is simple. I suspect that it can be derived avoiding Hermite polynomials, eigenvalues, etc. I understand them, but I think that short results are many times a consequence of short derivations.

    I found something, what can avoid this:
    http://ocw.mit.edu/courses/nuclear-...s-fall-2012/lecture-notes/MIT22_51F12_Ch9.pdf
    It seems that in pages 80 and 81 in sec. 9.1.2 is something, what maybe avoid this.

    For beginning, I do not understand, why
    ##N(a|n>)=(n-1)(a|n>)## gives that
    ##a|n=c_n|n-1>##
    Eigenvalue becomes eigenfunction.

    Probably this is an easy question, but I will help for further derivation.
     
    Last edited: Aug 2, 2016
  2. jcsd
  3. Aug 2, 2016 #2
    You're intuition is correct, there is certainly an easier way to determine the energy eigenvalues. The most common method is using the creation and annihilation operators (a and it's hermetian conjugate). If you want a good understanding of this method I would check out Griffiths quantum mechanics. He goes over both methods in chapter 2 I believe.
     
  4. Aug 2, 2016 #3

    BvU

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    is not correct. All it says is that ##a|n>## is an eigenvector of ##N##. Since its eigenvalue has been shown to be (n-1) it must be a multiple of ##|n-1>##.
     
  5. Aug 2, 2016 #4
    Still ever I do not understand, why it follows that it must be a multiple of ##|n-1>##? Is it possible to explain more clearly?
     
  6. Aug 2, 2016 #5

    Nugatory

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    If some given ket ##|\alpha\rangle## satisfies ##N|\alpha\rangle=(n-1)|\alpha\rangle## we know that that ket is an eigenket of ##N## with eigenvalue ##n-1##. The ket ##a|n\rangle## satisfies that equation, so we know that ##a|n\rangle## is an eigenket of ##N## with eigenvalue ##n-1##

    We also know that any eigenket of ##N## with eigenvalue ##n-1## can be written in the form ##c_n|n-1\rangle## with ##c_n## some constant. (You can substitute this into the eigenvalue equation to see that it works).

    Therefore we know that ##a|n\rangle## can be written as ##c_n|n-1\rangle##.
     
  7. Aug 3, 2016 #6
    This is, what I wished. Simple and short, without Hermite Polynomials.
     
  8. Aug 3, 2016 #7

    BvU

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    'Do' the Hermite polynomials at least once is my advice. Not all that complicated and you'll benefit a lot in later stages of your curriculum.
     
  9. Aug 3, 2016 #8
    Yes I calculated them and studied many aspects of them, but they are not an enough short answer, why relation ##E=\hbar\omega (n+1/2)## exists. This really need derivation in Griffiths. This derivation should be in and book with quantum oscillator.
     
  10. Aug 3, 2016 #9

    vanhees71

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    Everything in QM 1 (supposed you don't introduce spin) can be derived from the Heisenberg algebra of observables. For the harmonic oscillator you introduce creation and annihilation operators which fulfill the commutation relations
    $$[\hat{a},\hat{a}^{\dagger}]=1.$$
    The Hamiltonian reads
    $$\hat{H}=\hbar \omega \left (\frac{1}{2} + \hat{N} \right), \quad \hat{N}=\hat{a}^{\dagger} \hat{a}.$$
    Obviously ##\hat{N}## is a positive semidefinite operator, because for any ##|\psi \rangle##
    $$\langle \psi |\hat{N} \psi \rangle=\langle \psi|\hat{a}^{\dagger} \hat{a} \psi \rangle = \langle \hat{a} \psi|\hat{a} \psi \rangle \geq 0,$$
    because the scalar product is semidefinite.

    Now suppose ##|n \rangle## is an eigenvector of ##\hat{N}## with eigenvalue ##n \in \mathbb{R}##. Note that ##\hat{N}## is self-adjoint and thus can have only real eigenvalues. Now we have
    $$[\hat{N},\hat{a}]=[\hat{a}^{\dagger} \hat{a},\hat{a}]=[\hat{a}^{\dagger},\hat{a}]\hat{a}=-\hat{a}.$$
    This implies
    $$\hat{N} \hat{a} |n \rangle = ([\hat{N},\hat{a}]+\hat{a} \hat{N}) |n \rangle = (\hat{a} \hat{N}-\hat{a}) |n \rangle = (n-1) \hat{a} |n \rangle,$$
    i.e., ##\hat{a} |n \rangle## is either an eigenvector of ##\hat{N}## with eigenvalue ##n-1## or 0.

    This implies that applying ##\hat{a}^k## (##k \in \mathbb{N}##) to ##|n \rangle## you either get an eigenvector to the eigenvalue ##n-k## or ##0##. Since ##\hat{N}## is positive semidefinite, all eigenvalues are ##n \geq 0##. This implies that there must be a minimal eigenvalue ##n_{\text{min}}## such that ##\hat{N} |n_{\text{min}} \rangle=0##, but this implies that ##n_{\text{min}}=0##, because the former equation tells you that ##|n_{\text{min}} \rangle## is and eigenvector of ##\hat{N}## with eigenvalue 0.

    Now you have
    $$[\hat{N},\hat{a}^{\dagger}]=[\hat{a}^{\dagger} \hat{a},\hat{a}^{\dagger}]=\hat{a}^{\dagger} [\hat{a},\hat{a}^{\dagger}]=\hat{a}^{\dagger},$$
    and thus you find
    $$\hat{N} \hat{a}^{\dagger} |n \rangle = ([\hat{N},\hat{a}^{\dagger}]+\hat{a}^{\dagger})|n \rangle=(n+1) \hat{a}^{\dagger}|n \rangle,$$
    i.e., ##\hat{a}^{\dagger}## is an eigenvector of ##\hat{N}## with eigenvalue ##(n+1)##.

    Now you can build all eigenvectors from ##|0 \rangle## by repeated application of ##\hat{a}^{\dagger}##. It's easy to show with the commutation relations that the properly normalized eigenvectors of ##\hat{N}## are given by
    $$|n \rangle = \frac{1}{\sqrt{n!}} (\hat{a}^{\dagger})^n |0 \rangle.$$
    The eigenvalues of ##\hat{N}## are thus ##n \in \{0,1,2,3,\ldots \}=\mathbb{N}_0##, and thus the energy eigenvalues
    $$E_n=\hbar \omega \left (\frac{1}{2} +n \right), \quad n \in \mathbb{N}_0.$$
     
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