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Is the ground state of a harmonic oscillator unique?

  1. Feb 1, 2012 #1
    Hi,

    In one of my advanced quantum mechanics classes, the instructor posed a problem, namely to show that the ground state of a one dimensional quantum harmonic oscillator is unique, without getting into differential equations.

    I know that the equation

    [tex]a\left|0\right\rangle = 0[/tex]

    when written in the position space representation gives a simple differential equation, the solution to which is the familiar ground state Gaussian wavefunction. So, any such state which is a solution to the differential equation must be the same (up to a phase, which we can fix by a normalization choice).

    But how do you reason without using the differential equation approach? I was thinking about arriving at a proof by contradiction somehow, but it does not seem to me that the ground state being unique is a theorem (unless I am missing something more fundamental here), but rather a postulate.

    I digged into PF archives and found vanesch's statement that the ground state isn't in fact unique. The references are

    https://www.physicsforums.com/showthread.php?t=173896
    and
    https://www.physicsforums.com/showpost.php?p=1356434&postcount=2

    So, is the question wrong?
     
  2. jcsd
  3. Feb 1, 2012 #2
    The ground state is not unique. An example is the Landau level quantization generated by the Hamiltonian,

    [tex]H = \frac{(p-eA)^2}{2m}[/tex]

    This can be written as [itex]H = (a^\dagger a + \frac{1}{2})[/itex] with a suitable definition of the operator [itex]a[/itex]. So the energy spectrum is precisely that of the harmonic oscillator.

    But it turns out that for each energy level satisfying [itex]a^\dagger a|n\rangle n |\rangle[/itex] there are a macroscopic number of states. These degenerate energy levels are called Landau levels. In particular, the ground state is not unique.

    So it's very possible to have a number of orthogonal ground states [itex] |0_i\rangle [/itex]. Each of these ground states will generate a tower of harmonic oscillator states, [itex] a^\dagger|0_i\rangle[/itex]. You need a second quantum number to keep track of which tower you are in.

    You can get a unique ground state by demanding that the operators [itex]a[/itex] and [itex]a^\dagger[/itex] generate the entire Hilbert space when starting from any one state and you demand the existence of a lowest energy state.
     
  4. Feb 1, 2012 #3
    Interesting. I do not know enough of this to give you an answer, but I wanted to ask: Vanesch writes
    But I don't see how you have the freedom? Either the differential equation has multiple solutions for a certain energy level, or it doesn't. How can we afford to postulate anything about the number of solutions?
     
  5. Feb 1, 2012 #4

    martinbn

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    For the harmonic oscillator it is unique. You could 'cheat' and write down Hermite polynomials, check that they are eigenstates for the Hamiltonian and form an orthonormal basis of the state space. The last step needs a bit of Fourier analysis but no differential equations. Then the eigenstate with the smallest eigenvalue would obviously be unique. Well up to normalization.
     
  6. Feb 1, 2012 #5

    Demystifier

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    The two vacuums noted by vanesch are essentially the same, in the sense that these two vacuums generate two unitary equivalent irreducible representations of the oscillator algebra.

    One can prove that the vacuum is unique in the following sense. Assume that, besides the standard vacuum |0>, there is some additional vacuum |0'>. Since the standard set of N-eigenstates {|n>} is a complete basis, |0'> must have an expansion

    |0'> = c_0 |0> + Sum_{n=1}^{\infty} c_n |n>

    Act with the operator N on the left, and use N|0'>=0 (because |0'> is a vacuum). This gives

    Sum_{n=1}^{\infty} c_n n |n>=0

    which is only possible if c_n=0 for n=1,2,...,\infty. This implies

    |0'> = c_0 |0>

    i.e., the two vacuums differ only by multiplicative constant. Q.E.D.
     
  7. Feb 1, 2012 #6

    martinbn

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    But N is the number operator that has that property in the basis built from the vacuum |0>. Why should it be the same for the other basis? For example it is not the case in the context of the Unruh effect.
     
  8. Feb 1, 2012 #7
    Perhaps what's special about it is any level above it is simply a multiple of the ground state?
     
  9. Feb 1, 2012 #8
    That holds for any excited state.

    I agree, this is also how I thought about it. But there was an objection, namely that how can you expand |0'> in terms of the other basis? Does this not assume that |0'> generates the same space of states as |0> does when you operate the creation operator successively on it?

    I am familiar with the Landau level problem. However, the canonical momentum is not the same as the kinetic momentum in this problem, so even if the spectrum has the same form as a harmonic oscillator, this is physically a different problem: the magnetic field affects the ground state energy, where as in the free harmonic oscillator problem, the ground state energy is just the zero point energy which uniquely specifies the ground state.

    Also, the potential x^2 is different from switching on a constant magnetic field (as is the case in the Landau level problem) conceptually many other reasons...
     
  10. Feb 2, 2012 #9

    tom.stoer

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    The algebra of creation and annihilation operators alone is not sufficient to determine whether the ground state is unique; one has to specify in addition on which Hilbert space they are acting. Let's consider a simple example:

    [tex]A = \begin{pmatrix}a & 0 \\ 0 & a \end{pmatrix}[/tex]

    [tex]A^\dagger = \begin{pmatrix}a^\dagger & 0 \\ 0 & a^\dagger \end{pmatrix}[/tex]

    with the familiar algebra admit degenerate ground states

    [tex]|0\rangle_\theta = \begin{pmatrix}\cos\theta\;|0\rangle \\ \sin\theta\;|0\rangle \end{pmatrix}[/tex]
     
  11. Feb 2, 2012 #10

    Demystifier

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    I forgot to define the vacuum. What I meant by "vacuum" was any state |psi> satisfying
    a|psi>=0
    where "a" is a FIXED operator. If "a" was not fixed. i.e., if it was ANY operator satisfying the harmonic oscillator algebra, then you would be right.

    Recall that the exercise was to show that the vacuum IS unique. This, of course, cannot be shown without some additional assumptions.
     
    Last edited: Feb 2, 2012
  12. Feb 2, 2012 #11

    martinbn

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    Yes, I see.
     
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