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Time independant perturbation problem

  1. May 16, 2013 #1
    Example from Schaum's Quantum Mechanics. Picture of the example is attached.

    What I don't understand is part (c). What are those wavefunctions ##\mid \psi^{(0)}_{1,2} \rangle## and ##\mid \psi^{(0)}_{2,1} \rangle##? How do I find these wavefunctions, if the unperturbed wavefunction is ##\psi^{(0)}_{n_1,n_2}(x,y) = \frac{2}{L} \sin \left( \frac{\pi n_1 x}{L} \right) \sin \left( \frac{\pi n_2 y}{L} \right)##?
     

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  2. jcsd
  3. May 16, 2013 #2

    fzero

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    Those are not wavefunctions: they are quantum states. I am not familiar with Schaum, but the difference is discussed in any decent quantum text, such as Griffiths. Basically, in QM, the fundamental objects are the quantum states ##|\psi\rangle##. Given a Hamiltonian and other operators, ##\hat{A}##, that act on these states, we can define observables by acting on the states and computing things like ##\langle \psi | \hat{A} | \psi\rangle##.

    In this formalism, the position of a particle is also represented by an operator ##\hat{x}##. The eigenstates of this operator are quantum states we call ##|x\rangle##. The wavefunction corresponding to the state ##|\psi\rangle ## is defined by computing the inner product ##\psi(x) =\langle x | \psi \rangle##.

    Note that while we can often write down the wavefunction ##\psi(x)## as a function, the quantum state ##|\psi \rangle ## is often described more abstractly by specifying its quantum numbers, which are its eigenvalues with respect to a set of commuting operators.

    There are more details and references here.
     
  4. May 16, 2013 #3
    OK. So the quantum states given in part (c) are found by assigning ##n_1## and ##n_2## the values 1 and 2?
     
  5. May 16, 2013 #4

    fzero

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    In part. The quantum numbers ##n_1## and ##n_2## can be used to express the eigenvalue of the unperturbed Hamiltonian when acting on the states ##| \psi^{(0)}_{n_1,n_2}\rangle##. However, you want to compute the matrix element with the operator corresponding to the perturbation ##W##. In general (and in this particular case), the unperturbed states will not be eigenstates of the perturbation. So we actually want to express the matrix element in terms of the wavefunctions.

    We can do this by noting that the eigenstates ##|x\rangle## of the position operator form a complete set of states. The meaning of complete set of states might be discussed in your text, but the fact that we specifically need here is that the expression

    $$ \int d^2x |\vec{x}\rangle \langle \vec{x}| = 1$$

    as an operator statement (here ##\vec{x} = (x,y)##). So we can insert this in our matrix element

    $$ \langle \psi^{(0)}_{n_1,n_2} | W | \psi^{(0)}_{n_1,n_2}\rangle = \int d^2x \langle \psi^{(0)}_{n_1,n_2} | \vec{x}\rangle W\langle \vec{x} | \psi^{(0)}_{n_1,n_2}\rangle = \int d^2x (\psi^{(0)}_{n_1,n_2}(\vec{x}))^* W \psi^{(0)}_{n_1,n_2}(\vec{x}).$$

    The last term here gives us the matrix element in terms of the operator and the wavefunction. We can indeed determine a specific matrix element by picking the appropriate values of ##n_1## and ##n_2##, then computing the integral.
     
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