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Double asymmetric quantum well

  1. Feb 27, 2015 #1
    Hi,

    We know that for an infinite well of length a, eigenfunctions of the hamiltonian are :

    [tex]\psi_n(x)=\sqrt{\frac{2}{a}}sin(\frac{\pi n x}{a})[/tex] related to the eigenvalues [tex]E_n=n^2\frac{\hbar ^2 \pi^2}{2ma^2}[/tex]

    Now, I would like to consider two infinite quantum wells of length a et b (not necessarily equal) , like http://photonicssociety.org/newsletters/jun97/art/quantum3.gif [Broken].

    The eigenfunctions of the hamiltonian are clearly of the kind [tex]\psi_n^{(a_{well})}(x) + \psi_p^{(b_{well})}(x) [/tex] (n and p are not necessarily equal). But then, how could I find the (n,p) such that these are eigenfunctions. And then, what would be the eigenvalues ?

    I thank you in advance for your help.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Feb 27, 2015 #2

    Quantum Defect

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    The example shown are not infinitely deep.

    In finite double wells, you will see some intereting things.

    For symmetric double wells, you will see pairs of states:

    Psi_1+ = phi_1(a) + phi_1(b)
    Psi_1- = phi_1(a) - phi_1(b)

    The energies of these states are close to the energies of the phi_1(a/b) states. The states are said to be "split by tunneling." You will see pairs of these all theway up to the top. As the barrier between the wells gets lower, the tunneling splitting increases.

    For asymmetric double wells, like your example, you will see states that look to be slightly shifted versions of the "basis states." The lowest state in your example, looks like n=1 for the wide well., while the next two look like linear combinations of n=1 in the narrrow well and n=2 in the wide well. The mixing coefficients are not equal, as is the case in the tunneling split levels of the symmetric case. E2 is probably closer to the unperturbed n=2 level of the wide well, and E3 is probably closest in energy to the unperturbed n=1 level of the narrow well.

    You will need to calculate these wavefunctions numerically, just as you would for the states in the finite well. [i.e. match wavefunctions and first derivatives at key points]. There is an old paper by Cooley that discusses an efficient way to calculate the eigenfunctiond and eigen energies of one-dimensional potentials. [So-called Cooley-Numerov method.] See e.g. http://www.myoops.org/cocw/mit/NR/rdonlyres/Chemistry/5-73Introductory-Quantum-Mechanics-IFall2002/2139DA2B-09EC-4A27-A89D-0FF4666D5B13/0/notes09.pdf [Broken] for a discussion of the Cooley-Numerov method. This presentation also gives the reference to the original paper by Cooley.
     
    Last edited by a moderator: May 7, 2017
  4. Feb 27, 2015 #3
    Thanks for your answer ! Now I understand it.
     
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