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About momentum eigenfunctions

  1. Jun 11, 2010 #1
    I'm enjoying this introductory essay about quantum mechanics found here


    and I have a question. About five-eighths of the way into it a wave function is given "at time t=0",


    \psi = \sqrt { \frac {2} {L} } [ \frac {1}{2} sin (\pi x / L) + \frac { \sqrt {3} }{2} i sin (5 \pi x / L)]


    and some questions and answers follow. If I am understanding the authors, the answers imply that this wavefunction is a (normalized) linear combination of two momentum eigenfunctions, where the momenta are [itex] h/2L [/itex] and [itex] 5h/2L [/itex].

    My question is, shouldn't


    \psi = \sqrt { \frac {2} {L} } [ \frac {1}{2} (cos (\pi x / L) + i sin (\pi x / L)) + \frac { \sqrt {3} }{2} (cos (5 \pi x / L)
    + i sin (5 \pi x / L))]


    or - more succinctly -


    \psi = \sqrt {\frac {2} {L} } [ \frac {1}{2} e ^ {i \pi x / L } + \frac { \sqrt {3} } {2} e ^ {i 5 \pi x / L } ]


  2. jcsd
  3. Jun 11, 2010 #2
    Oops - I just looked backward and noticed that this is one of those infinite potential well situations. That explains why both the real and imaginary components of [itex] \psi [/itex] have to be sine waves here.

    And yet, how do I square this fact with the postulate that

    e ^ { ipx/ \hbar }

    - not [itex] sin (px / \hbar ) [/itex] - are the momentum basis states?
  4. Jun 11, 2010 #3
    Write the sin as exp(ipx) - exp(-ipx), so isnt the sine just a different basis?
  5. Jun 11, 2010 #4
    No, the answers imply that this wavefunction is a (normalized) linear combination of two ENERGY eigenfunctions, not momentum.
  6. Jun 11, 2010 #5
    Ah, yes - that did it. Thanks sonoluminated.

    When the potential energy is zero, don't those amount to the same thing?
  7. Jun 11, 2010 #6
    Last edited by a moderator: Apr 25, 2017
  8. Jun 11, 2010 #7
    That's a nice source, thanks sweet springs.

    What I meant was, since inside the well the potential energy is zero, there is a one-to-one relation between the discrete momentum states and the discrete energy states.
  9. Jun 11, 2010 #8
    Hi, snoopies622.
    Shubert (3.22),(3.23) and Figure 3.3 show us that the discrete energy eigenstate is superposition of continuous momentum eigenstates.
    Energy at somewhere e.g. inside the well means simultaneous consideration of energy E or Hamiltonian H and position X. Commutation relation [H,X] ≠0 does not allow such consideration.
    Last edited: Jun 11, 2010
  10. Jun 12, 2010 #9
    I haven't studied the math yet but conceptually I am confused. In a place where there is no force acting on a particle and the potential energy is zero, the total energy must be the kinetic energy, which is a function of the momentum. How then can discrete energy states not imply discrete momentum states, and vice versa?
  11. Jun 12, 2010 #10
    Hi, snoopies622.
    I say here not using mathematics so do not take it so strict. See ψ0 in Shubert Figure 3.2. that is the ground energy state. You see near the well walls particle rarely exists and around center of the well particle is most probably found. This behavior is different from classic particle whose probability density is uniform in the well. The particle seems to be knowing the distances from the walls though local information e.g. potential energy does not change as you stated;
    In quantum physics such a global behavior replaces classical localism. In quantum physics amount of energy cannot be distinguished to kinetic and potential parts according to where the particle is.
    Last edited: Jun 12, 2010
  12. Jun 13, 2010 #11
    By "in a place" I didn't mean at a particular location inside the well, just inside the well. That is, at any location inside the well, the potential energy is zero and therefore all the energy is kinetic energy. I think we agree on this.
  13. Jun 13, 2010 #12
    Hi, snoopies622
    In quantum mechanics, even if potential energy is zero anywhere inside the well, energy eigenstate is under the influence of the potential beyond the walls. I called it "global" in my previous post. Tunnel effect is on the same footing.
    Last edited: Jun 13, 2010
  14. Jun 14, 2010 #13
    The problem is that [itex] \psi [/itex] as described above


    \psi = \sqrt { \frac {2} {L} } [ \frac {1}{2} sin (\pi x / L) + \frac { \sqrt {3} }{2} i sin (5 \pi x / L)]


    can be expressed equivalently as


    \psi = \sqrt { \frac {2} {L} } [ \frac {i}{4} (e ^ {-i \pi x / L} - e ^ {i \pi x / L}) + \frac { \sqrt {3} }{4} (e ^ {5i \pi x / L} - e ^ {-5i \pi x / L})]


    which is a linear combination of four momentum eigenfunctions,


    e ^ {-i \pi x / L} \quad
    e ^ {i \pi x / L } \quad
    e ^ {5i \pi x / L} \quad
    e ^ {-5i \pi x / L}

    not a continuum of them.
  15. Jun 14, 2010 #14
    Hi. Snoopoies622
    You are right in case the formula stands from -infinity x to +infinity x. I am afraid you are not right in case the formula stands only within the well and turns to zero for the outside. Which one is your case?
    Last edited: Jun 14, 2010
  16. Jun 14, 2010 #15
    Because when the particle is in an external potential, the momentum does not commute with the total Hamiltonian of the particle and, hence, it does not have a definite value in a stationary state.
  17. Jun 14, 2010 #16
    I was wondering about that...

    So when the energy of the particle (inside the well) is measured to be [itex] E _n [/itex] we cannot assume that its momentum is therefore [tex]

    \pm \sqrt {2m E_n }

    [/tex] ?
  18. Jun 14, 2010 #17
    Hi. snoopies622
    No, in case the system has potential energy so there is a well.
    Yes, in case the system has no potential energy so there is no well.
    Last edited: Jun 14, 2010
  19. Jun 14, 2010 #18
    This surprises me. I always looked at the particle-in-a-one-dimensional-box situation like this:

    The wave must fit neatly inside the box, so the wavelength must be 2L/n where n is integer. Therefore, the particle's momentum must be


    p = \frac {h}{\lambda} = \frac {hn}{2L}


    and its energy


    E_n = \frac {p^2}{2m} = \frac {h^2 n^2}{4 L^2} \frac {1}{2m} = \frac {\hbar ^2 \pi ^2 }{2 m L^2} n^2

  20. Jun 15, 2010 #19
    Hi, snoopies622.
    Fig.1 is wave function of momentum eigenstates p=h'/λ and p=-h'/λ superposed in equal weight.
    Fig.2 is wave function of an energy eigenstate in infinite square well potential.
    Though Fig.1 and Fig.2 coincide within the well region, They are different states. Fourier transform of Fig.2 gives continuous momentum spectrum. See Schubert.
    You can rewrite En as pn^2/2m where pn is a quantity whose dimension is momentum, but pn is not momentum of the system. However as classical limit taking n large, the state momentum tends to +-pn. See Schubert Fig3.2.
    Last edited: Jun 15, 2010
  21. Jun 16, 2010 #20
    I think I understand now. I have one more question: what does the Schrodinger equation look like in the momentum basis (instead of the position basis)?
  22. Jun 16, 2010 #21
    Hi, snoopies622.
    Shrodinger equations of stationary state for Hamiltonian H(p,x) are H(h'/i d/dx, x) ψ(x)=Eψ(x) in coordinate space and H(p, ih'd/dp) φ(p)=Eφ(p) in momentum space where φ(p) is Fourier transform of ψ(x).
    Last edited: Jun 16, 2010
  23. Jun 16, 2010 #22
    Thanks for that, sweet springs.

    I'm still having a problem with the mathematical equivalence I mentioned in entry 13. I see how neither representation takes into account the boundaires of x=0 and x=L, and that this is why the momentum representation (the second equation) doesn't work. But why then does the first one? I assume that a function that is sinusoidal between two x values and then suddenly zero outside of that domain is much more complicated than the sum of either two or four sinusoidal terms.
  24. Jun 16, 2010 #23
    Schroedinger Equation in momentum basis is:

    \frac{p^{2}}{2 m} \, a(\mathbf{p}) + \int{\frac{d^{3}p}{(2 \pi \hbar)^{3}} \, \tilde{U}(\mathbf{p} - \mathbf{p}') \, a(\mathbf{p}') = E \, a(\mathbf{p})


    a(\mathbf{p}) \equiv \int{d^3{r} \, \psi(\mathbf{r}) e^{-\frac{i}{\hbar} \mathbf{p} \cdot \mathbf{r}}}

    is the wave function in momentum space. It's physical significance is that:

    |a(\mathbf{p})|^{2} \frac{d^{3}p}{(2 \pi \hbar)^{3}}

    is the probability that the particle will have a value of momentum inside an elementary cube of volume [itex]d^{3}p[/itex] around the value [itex]\mathbf{p}[/itex], and

    \tilde{U}(\mathbf{p}) = \int{d^{3}r \, U(\mathbf{r}) \, e^{-\frac{i}{\hbar} \mathbf{p} \cdot \mathbf{r}}}

    is the spatial Fourier transform of the potential energy.

    If the potential depends only on one Cartesian coordinate (denoted by x), then:
    \tilde{U}(\mathbf{p}) = \tilde{U}(p_{x}) \, (2 \pi \hbar)^{2} \, \delta(p_{y}) \, \delta(p_{z})

    and we can choose a wave-function with a definite value of the components of the momenta in the y and z direction:

    a(\mathbf{p}) = a(p_{x}) \, (2 \pi \hbar)^{2} \, \delta(p_{y} - p_{y0}) \, \delta(p_{z} - p_{z0})


    |a(p_{x})|^{2} \frac{dp_{x}}{2 \pi \hbar}

    is the probability that [itex]p_{x}[/itex] has a value in the interval [itex](p_{x}, p_{x} + dp_{x})[/itex]. The Schroedinger equation becomes one dimensional:

    \frac{p_{x}^{2}}{2 m} \, a(p_{x}) + \int{\frac{dp_{x}}{2 \pi \hbar} \, \tilde{U}(p_{x} - p'_{x}) \, a(p'_{x})} = \epsilon \, a(p_{x}), \ \epsilon = E - \frac{p^{2}_{y} + p^{2}_{z}}{2 m}
  25. Jun 17, 2010 #24
    Hi. snoopies622
    Both the representation take into account the potential energy generating boundaries at x=0 and x=L. The coordinate representation is easier to solve because equation includes up to (d/dx)^2. In the momentum representation equation (d/dp)^n of any large n appear corresponding to power series expression of V(x). Dickfore discussed it in another way using Fourier transform of the potential energy.
  26. Jun 17, 2010 #25
    Wow, thanks Dickfore. I'll have to give all that some time.

    Actually, the two representations I had in mind in my previous question were not position vs. momentum. They were both in position representation. (See entry #13.) The first was a linear combination of energy eigenfunctions, and the second was a linear combination of momentum eigenfunctions. I was wondering why one could use the first one to find the particle's allowed energy values, but not the second one to find its allowed momentum values.
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