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Infinite potential well

  1. Feb 1, 2013 #1
    In one dimensional problem of infinite square potential well wave function is ##\phi_n(x)=\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L}## and energy is ##E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}##. Questions: What condition implies that motion is one dimensional. Did wave function describes motion of particle? When we say particle did we mean electron? What kind of particle mass are good enough to be treated in this model? And also why we don't draw ##\phi_n(x)##.
     
    Last edited: Feb 1, 2013
  2. jcsd
  3. Feb 2, 2013 #2
    one distance variable x, implies one dimension.


    http://en.wikipedia.org/wiki/Schrödinger_equation

    be sure to see the dynamic illustration near the beginning of the article here....http://en.wikipedia.org/wiki/Infinite_square_well
     
  4. Feb 2, 2013 #3
    The picture you must draw is a string (one dimension) vibrating between two fixed point (the walls of the potential) - distance L. The harmonics are the quantum excitations of the same particle or elementary system - p_n = n h / \lambda . This gives you the quantized momentum spectrum, that is the harmonic discrete spectrum a guitar string. The energy spectrum follows from the non-relativistic dispersion relation E = p^2 / 2 m.
     
  5. Feb 2, 2013 #4
    I didn't get the answer that I'm looking. My questions:
    Did wave function describes motion of particle?

    What kind of particle mass are good enough to be treated in this model?

    And

    And also why we don't draw ##\phi_{-n}(x)##?
     
  6. Feb 3, 2013 #5
    For motion you need to solve the Schrodinger equation. The wavefunctions you are talking about are solutions to the energy eigenvalue equation, but it is the Schrodinger equation that gives you the evolution of state in time.

    Anything I guess. You have an expression involving m, and you can plug in whatever you want for m. If m is large then the energy levels are closely spaced which is good as that is common experience.


    [itex]\phi_{-n}(x)[/itex] is the same state as [itex]\phi_{n}(x)[/itex] as they only differ by a phase.
     
  7. Feb 3, 2013 #6
    What about post #2 did you not understand?

    Did you read the links I posted.....they will help a lot.

    " Interpretations of quantum mechanics address questions such as what the relation is between the wavefunction, the underlying reality, and the results of experimental measurements."

    There are different interpretations about exactly what the wavefunctions means...there are different ways to think about it.
     
  8. Feb 3, 2013 #7
    Usually in one dimensional problem you can constrain a particle between two infinitely large parallel plates, that makes the particle only subject to the boundary conditions in one dimension.

    Also, I don't think that particles are described by equations of motion in QM. In QM we describe particles in Hilbert space, in classical mechanics we use phase space which describe the equations of motion of a particle.

    Basically any particles follows QM, but when n is large, by Correspondence Principle, classical mechanics must be a good approximation of QM
     
  9. Feb 5, 2013 #8
    One more question. By solving Sroedinger eq we get
    ##\varphi_n(x)=\sqrt{\frac{2}{a}}\sin \frac{n\pi x}{a}##
    ##E_n=\frac{n^2\pi^2\hbar^2}{2ma^2}##
    But for example solution of Sroedinger eq is also
    ##C_1sin\frac {\pi x}{a}+C_2\sin\frac{2\pi x}{a}##
    in which state is particle?
    What is the energy level? How could I know in which state is particle?
     
  10. Feb 5, 2013 #9

    jtbell

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    Staff: Mentor

    The particle is in a superposition of the n=1 and n=2 states. When something happens that in effect measures the particle's energy, it will be either E1 or E2, chosen randomly according to probabilities that depend on C1 and C2.

    Before the measurement happens, the answer to the question, "which of the two energy states is the particle really in?" depends on which interpretation of QM you subscribe to.
     
  11. Feb 5, 2013 #10
    Well from
    [tex]\int^{a}_0 (C_1\sin \frac {\pi x}{a}+C_2 \sin \frac{2\pi x}{a})^2=1[/tex]
    I get
    ##C_1^2+C_2^2=\frac{2}{a}##
    what next?
     
  12. Feb 5, 2013 #11

    jtbell

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    Staff: Mentor

    It's better to start out by normalizing each individual function:

    $$\psi(x) = C_1 \psi_1(x) + C_2 \psi_2(x)\\
    \psi(x) = C_1 \sqrt{\frac{2}{a}} \sin {\left(\frac{\pi x}{a}\right)}
    + C_2 \sqrt{\frac{2}{a}} \sin {\left(\frac{2 \pi x}{a}\right)}$$

    Then when you normalize ##\psi## as a whole by doing an integral like yours, you end up with

    $$C_1^2 + C_2^2 = 1$$

    We interpret ##C_1^2## as the probability that the particle will be measured to have energy E1, and ##C_2^2## as the probability that the particle will be measured to have energy E2. The two probabilities add to 1 because they are the only two possibilities for this particle as far as energy is concerned.
     
  13. Feb 6, 2013 #12
    Ok but from ##C_1^2+C_2^2=1## and some other physical behavior could I say something about ##C_1^2## and ##C_2^2##?
     
  14. Feb 6, 2013 #13

    jtbell

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    Staff: Mentor

    Obviously you need additional information about your specific situation in order to find C1 and C2. It's impossible to say more unless you have a specific situation in mind.
     
  15. Feb 7, 2013 #14
    I understand that. But could you give me example of some specific situation.
     
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