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B Ladder operator

  1. Oct 9, 2016 #1
    We learnt that we can use the ladder operator to obtain the states of a quantum oscillator. However, I see no direct evidence to show that the solutions are complete. I mean, how can we know the energy state follows E is (E+hw). Why can't we have some more states in between? Does the derivation using ladder operator prove this?
     
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  3. Oct 10, 2016 #2

    Mentz114

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    Have a look at the Wiki article, which answers your questions.

    https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator
     
  4. Oct 10, 2016 #3

    stevendaryl

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    As is said in the Wikipedia article Mentz114 mentioned, suppose you have a state [itex]|\psi\rangle[/itex] with positive energy eigenvalue [itex]E[/itex]. If you apply the lowering operator [itex]a[/itex], you'll get another eigenstate with energy [itex]E-\hbar \omega[/itex]. If you keep applying [itex]a[/itex], you will come up with a state [itex]|\psi'\rangle \equiv a^n |\psi\rangle[/itex] such that [itex]H |\psi'\rangle = (E - n \hbar \omega) |\psi'\rangle[/itex]. Pick [itex]n[/itex] large enough so that [itex]E - n \hbar \omega < 0[/itex]. It's impossible for the energy to be negative (for this particular hamiltonian, anyway). So that's a contradiction, unless [itex]|\psi'\rangle = 0[/itex]. So we conclude that:

    If [itex]|\psi\rangle[/itex] is any energy eigenstate, then there is some number [itex]n[/itex] such that [itex]a^n |\psi\rangle = 0[/itex]
     
  5. Oct 10, 2016 #4

    vanhees71

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    Additionally you have to assume that the representation of the usual operator algebra should be irreducible. Then the harmonic oscillator Hamiltonian is non-degenerate (for this one-dimensional case), and you get a complete set of orthogonal eigenstates, because the Hamiltonian is self-adjoint.
     
  6. Oct 10, 2016 #5
    Well, it follows from the Schrödinger equation for the harmonic oscillator. If you solve (e.g. by power series method) it and apply the boundary conditions and normalization, you find that the energy is quantized, ##E_n=\left(n+\frac{1}{2}\right)\hbar \omega##.
     
    Last edited: Oct 10, 2016
  7. Oct 10, 2016 #6

    stevendaryl

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    I interpreted the question as asking how to demonstrate this using the ladder operators.
     
  8. Oct 10, 2016 #7
    I think that in order to use the formalism of the ladder operators you have to presuppose that the oscillator is quantized.
     
    Last edited: Oct 10, 2016
  9. Oct 10, 2016 #8

    vanhees71

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    No, you use the ladder operators to derive that the energy of the ho is quantized. Since ##\hat{H}## is positive semidefinite and ##\hat{a}##, applied to any energy eigenstate leads to an energy eigenstate with an energy by ##\hbar \omega## lower or 0. Thus there must be an energy eigenstate ##|\Omega \rangle## which is mapped to 0 by ##\hat{a}##. Then using ##\hat{a}^{\dagger}## you can get all energy eigenstates by applying it multiple times to ##|\Omega \rangle##. Assuming then that the Heisenberg algebra is realized irreducibly this leads to a complete orthnormal basis of energy eigenvectors.
     
  10. Oct 10, 2016 #9

    stevendaryl

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    No, I would say that the ladder operators prove this, rather than presuppose it.
    1. All eigenvalues of [itex]H[/itex] are positive.
    2. For any eigenstate [itex]|\psi\rangle[/itex] of [itex]H[/itex], if [itex]H|\psi\rangle = E |\psi\rangle[/itex], then for any positive integer [itex]n[/itex], [itex]H (a^n |\psi\rangle) = (E - n \hbar \omega) (a^n |\psi\rangle)[/itex].
    3. If [itex]E - n \hbar \omega < 0[/itex], this is only possible if [itex]a^n |\psi\rangle = 0[/itex]
    4. So let [itex]n[/itex] be the smallest non-negative integer such that [itex]a^{n+1} |\psi\rangle = 0[/itex].
    5. Then the state [itex]|\psi_0\rangle \propto a^n |\psi\rangle[/itex] is nonzero, and satisfies [itex]a |\psi_0\rangle = 0[/itex]. (I'm using [itex]\propto[/itex] to mean equal, up to a multiplicative normalization constant).
    6. There is only one normalizable state satisfying that equation.
    7. If [itex]a^n |\psi\rangle \propto |\psi_0\rangle[/itex], then we can show that [itex](a^\dagger)^n |\psi_0\rangle \propto |\psi\rangle[/itex]
    8. So since [itex]|\psi_0\rangle[/itex] is unique, that shows that every eigenstate [itex]|\psi\rangle[/itex] can be written in the form [itex]|\psi\rangle \propto (a^\dagger)^n |\psi_0\rangle[/itex] for some nonnegative integer [itex]n[/itex].
    9. So the eigenstates are quantized.
     
  11. Oct 10, 2016 #10
    I think you presuppose that ##E=m\hbar\omega+c##, where ##m## is integer and ##c<\hbar\omega##.
    In this case there is indeed a unique ##\psi_0## with energy ##E_0=c##, and any other eigenstate ##\psi## is created by applying ##\hat{a}^n## on ##\psi_0##.
     
    Last edited: Oct 10, 2016
  12. Oct 10, 2016 #11

    stevendaryl

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    But if [itex]E[/itex] is a real number, then it's got to be the case that [itex]E[/itex] can be written as [itex]E = n \hbar \omega + c[/itex] for some [itex]c \leq \hbar[/itex].
     
    Last edited: Oct 10, 2016
  13. Oct 10, 2016 #12
    Not if ##c## is the same for all ##E##, as you have supposed in your proof.
    In other words, why in your proof ##\psi_0## is unique (same for any ##\psi##)?
     
  14. Oct 10, 2016 #13

    stevendaryl

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    Because there is only one solution to the equation [itex]a |\psi_0\rangle = 0[/itex] (up to a multiplicative constant).

    If you look at what [itex]a[/itex] is, it's

    [itex]a = \sqrt{\frac{m \omega}{2 \hbar}} (x + \frac{i}{m \omega} p) = \sqrt{\frac{m \omega}{2 \hbar}} (x + \frac{\hbar}{m \omega} \frac{\partial}{\partial x})[/itex]

    So [itex]a |\psi_0\rangle = 0[/itex] means [itex]x \psi_0(x) + \frac{\hbar}{m \omega} \frac{\partial}{\partial x} \psi_0(x) = 0[/itex]

    So [itex]\frac{\partial \psi_0}{\partial x} \frac{1}{\psi_0} = -\frac{m \omega x}{\hbar}[/itex]

    The left-hand side is just [itex]\frac{\partial log(\psi_0)}{\partial x}[/itex]. So we have:

    [itex]\frac{\partial log(\psi_0)}{\partial x} = - \frac{m \omega x}{\hbar}[/itex]

    Integrate both sides with respect to [itex]x[/itex] to get:

    [itex]log(\psi_0) = - \frac{m \omega x^2}{2 \hbar} + C[/itex]

    where [itex]C[/itex] is some constant. So:

    [itex]\psi_0 = e^{- \frac{m \omega x^2}{2 \hbar} + C} = C' e^{- \frac{m \omega x^2}{2 \hbar}}[/itex]

    where [itex]C' = e^{C}[/itex]. So [itex]\psi_0[/itex] is uniquely determined (up to the multiplicative constant [itex]C'[/itex])

    I don't know of a way to prove that [itex]a |\psi_0\rangle = 0[/itex] has a unique solution without actually looking at what [itex]a[/itex] is.
     
  15. Oct 11, 2016 #14

    vanhees71

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    Well, this however assumes that the ground state of the HO is non-degenerate. Take a particle with spin with the HO potential only (i.e., no magnetic fields around). Then the ground state (and thus all other energy eigenstates) are ##(2s+1)##-fold degenerate.
     
  16. Oct 12, 2016 #15
    All of your views give me new insight into this! Thanks! But I am still digesting your words.....
     
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