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Purely imaginary bound state wave functions?

  1. Sep 19, 2008 #1
    Hello all. I’m researching rotational motion with a nearly harmonic potential using the basis of the particle on a ring eigenstates e(n*i*theta) defined from theta=0 to theta=2*pi.
    The total systems wave functions (eigenfunctions of the full Hamiltonian (KE+PE)) are then linear combinations of the particle on a ring eigenstates.

    The thing that has me stumped is this: As the potential is symmetric around pi, the wave functions are necessarily either symmetric or anti-symmetric around pi. If you describe an anti-symmetric wave function with these types of states however, you’ll see that the real part cancels out and you’re left with all of the amplitude due to the imaginary part alone.
    That is, for a given value of n, the coefficient in front of e(n*i*theta) is either equal to that in front of the e(-n*i*theta) term (then the i*sin(theta) part cancels because sin is an odd function) or is equal to -1*that coefficient (and the cosine part cancels because itis an even function).


    I was under the impression that bound state wave functions MUST be real. However, I don’t see how anti-symmetric wave functions can exist in this system if they’re NOT purely imaginary and I don’t know of any reason to exclude the anti-symmetric states apart from that.

    By “I was under the impression that…” I mean that it’s something that I’ve heard on the street and never really thought about. Can someone either help me out with how I’m looking at this incorrectly or tell me why “all bound states must be purely real” is a hard fast rule of quantum mechanics and I should exclude the asymmetric states from my analysis. This seems strange because asymmetric states of the harmonic oscillator aren't excluded and this system is basically the same but with a weird basis.

    Any help would be much appreciated. Thanks in advance.
     
  2. jcsd
  3. Sep 19, 2008 #2

    olgranpappy

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    Bound state wave functions, like any eigenfunctions of the hamiltonian, don't *have to* be real--but may be *chosen* as real.

    For example, the hydrogen atom; the 1s and 2s wavefunctions are real, the 2p wavefunctions are proportional to the spherical harmonics [tex]Y_{1m}[/tex] which aren't always real. Of course, you can chose linear combinations of 2p orbitals that are real and work with those.
     
    Last edited: Sep 19, 2008
  4. Sep 19, 2008 #3

    Hurkyl

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    More specifically... What happens if you take the antisymmetric solution, and let the coefficient by purely imaginary, say (1/2i)?
     
  5. Sep 19, 2008 #4

    atyy

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    The wavefunction of a free particle ~exp(ikx), so that's probably what you're thinking a bound state wavefunction shouldn't be.

    The global phase of a wavefunction doesn't matter, and we can multiply all wavefunctions by exp(i.theta), because we only use |wavefunction|2 to make predictions. However, the relative phases of wavefunctions matter since |A+exp(i.theta)B|2 is not the same as |exp(i.theta)(A+B)|2.
     
  6. Sep 19, 2008 #5
    You're doing the same thing people do when they swith between what I call the physicists basis states and the chemist's basis states for the hydrogen atom. You know when they show pictures of the hydrogen orbitals in chemistry textbooks, how they have 3 dumbells along the x,y, and z axes for the p states? That comes from taking sums and differences of the l=+/-1 orbitals. So you would get a pure real and pure imaginary state, like you have for your ring states. (This idea of sums and differences also explains the funny shapes they get when they draw the d orbitals).
     
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