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In an infinite quantum well, why Δn=0?

  1. Feb 2, 2016 #1
    I've been reading up a bit on semiconductor quantum wells, and came across a selection rule for an infinite quantum well that says that "Δn = n' - n = 0", where n' is the quantum well index of an excited electron state in the conduction band, and n is the index of the valence band state where the electron came from (see, for example, p. 126 of "Optical Properties of Solids" by Mark Fox (2001)). Let's assume that I'm operating within all the basic approximations (e.g., we're using light to excite this electron, the quantum well is oriented parallel to x and y, and perpendicular to z, the dipole approximation is valid such that the interaction Hamiltonian is given by E*x, light is incident normal to the quantum well such there is no component of E oriented normal to the quantum well, screening and Coulomb interactions are ignored, etc.).

    It is usually the case that a relevant conservation law must exist because of some underlying symmetry. For example, I can understand that the selection rule "Δn ∈ evens" is a consequence of parity conservation, which is in turn a consequence of inversion symmetry.

    My question is this: Is there an underlying symmetry in the Hamiltonian forcing the selection rule "Δn = 0" to exist? What is it?

    I suppose my best guess is that it has something to do with momentum conservation, although it's a little hard to wrap my head around this since the quantity conserved is not "k" in the usual sense, but rather "k_nz = n*pi/w", (where "n" is a positive integer and "w" is the well width), and there is no translation symmetry along the "z" direction. Maybe there's a symmetry related to continuous transformations of the well width?
  2. jcsd
  3. Feb 2, 2016 #2


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    I don’t think so. The symmetry of an infinite quantum well is exactly the same as the symmetry of a finite quantum well but the selection rule for a finite quantum well is Δn=even. It’s true that symmetries of the Hamiltonian lead to conservation laws, but it’s the symmetry of the eigenvectors of the Hamiltonian which determine selection rules.

    In the case of an infinite quantum well you can write the wavefunction of the electrons (or holes) as a Bloch part multiplied by an envelope function, where the envelope function is just the solution of the time-independent Schrodinger equation for the well potential. For the case of an infinite well, the form is well-known and simple: Ψenvelope=sin(z*k_nz).

    Clearly, the envelope functions are orthogonal for different values of n so the dipole matrix element vanishes for n’ not equal to n.

    In the case of a finite quantum well, Ψenvelope extends into the barrier and in this case, it you can see that the overlap integral will only vanish if n’ is odd and n is even. (Or vice versa)
  4. Feb 2, 2016 #3
    Ah! My reference text misled me. It seems to be the case that Δn = 0 must hold even for the finite square well.

    Consider two bound states |1> and |2>, with distinct energies E1 and E2, and consider the matrix element,

    [itex]\langle 1 | H | 2\rangle[/itex].

    H can operate to either the left or the right. Thus,

    [itex]E1 \langle 1 | 2\rangle = E2 \langle 1 | 2\rangle[/itex],

    which can only be true E1 = E2 (which is false by construction), or if

    [itex]\langle 1 | 2\rangle = 0[/itex].

  5. Feb 3, 2016 #4


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    For many real quantum wells, such as GaAs/AlGaAs wider than about 100Angstroms, Δn=0 is a reasonable approximation because the 2D-confinement is quite strong. But it is not a selection rule and the other transitions, though weaker, are not forbidden. Only if the well is infinite are they strictly forbidden.

    I just did a little google search and found conflicting information (probably as a result of the above.) Here are some results that get it right:

    (See in particular the second paragraph of page 6: “This rule is somewhat weaker for finite wells, although it is still a very good starting point.” This also contains a discussion of the selection rules in terms of momentum conservation.)

    (Page 4)

    And here’s a more thorough treatment:
    (See section 10.4.2)
  6. Feb 3, 2016 #5
    Thanks, this ought to help.
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