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Time evolution of a quantum system

  1. Nov 25, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Let the time evolution of a system be determined by the following Hamiltonian: $$\hat{H} = \gamma B \hat{L}_y$$ and let the system at t=0 be described by the wave function ##\psi(x,y,z) = D \exp(-r/a)x,## where ##r## is the distance from the origin in spherical polars. Find the state of the system at any time t. Deduce the expectation value of ##\hat{L}_z## in the state ##\psi##.

    2. Relevant equations
    In Cartesians, $$\hat{L}_y = -i\hbar \left(z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z}\right)$$

    3. The attempt at a solution
    Computing the Hamiltonian gives the new wave function ##-Di \hbar \gamma B z \exp(-r/a)##. To obtain the time dependence, by solving the spatial-independent Schrodinger equation, $$\psi(x,y,z,t) = \psi(x,y,z)\exp(-Et/\hbar).$$ I am not sure if this is correct since ##\psi(x,y,z)## is not an eigenstate of ##\hat{L}_y## and hence not of ##\hat{H}## either.
    Thanks.
     
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  3. Nov 25, 2013 #2

    vela

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    You're right. You can't use that expression for the time evolution of a generic state. It only works for eigenstates.

    You need to express the initial wave function as a linear combination of the eigenstates of the Hamiltonian because you know how to write down the time evolution of an eigenstate. So the first thing you need to do is figure out what those eigenstates are.
     
  4. Nov 25, 2013 #3

    CAF123

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    I forgot to mention in a previous part of the question, we are given eigenfunctions of ##\hat{L}_z## and ##\hat{L}_y##. For ##\hat{L}_y## they are ##\phi (x,y,z) = f(r)(z\pm ix)##, corresponding to eigenvalues ##\pm \hbar##. So these are also the eigenfunctions of ##\hat{H}## with e.values ##\pm \gamma B \hbar##. Hence, $$\psi (x,y,z) = \sum_{n=1}^2 c_n \phi (x,y,z) = c_1 f(r) (z+ix) + c_2 f(r) (z-ix)$$ and the ##c_i## are obtained by projecting the wave function ##\psi## onto this basis: $$c_n = \langle \psi | \phi\rangle = \int_{\text{all space}} x \exp(-r/a)f(r)(z+ix)\,d\tau$$ and similarly for the other coefficient. Am I supposed to do this integral?
     
  5. Nov 25, 2013 #4

    vela

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    That would be one way, but there's a simpler way. Remember the goal is to express ##\psi## in terms of the eigenfunctions. You want to choose ##c_1## and ##c_2## such that
    $$c_1 f(r) (z+ix) + c_2 f(r) (z-ix) = D e^{-r/a}x$$ holds for all x, y, and z. You should be able to see what f(r) is by inspection. Collect terms on the lefthand side and then match coefficients for x and z between the two sides to get a system of equations for ##c_1## and ##c_2##.
     
  6. Nov 25, 2013 #5

    CAF123

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    I see, so in the end, $$\psi (x,y,z,t) = \frac{D}{2} f(r) (z+ix) \exp(-iE_1t/\hbar) - \frac{D}{2} f(r) (z-ix)\exp(-iE_2t/\hbar)$$ where ##E_1## and ##E_2## are the energy eigenvalues corresponding to the each function in the basis.

    To compute the expectation value of ##\hat{L}_z## in the state ##\psi## does that mean deal with the wave function at time t=0?, so compute ##\langle \hat{L}_z \rangle = \int \psi^* \hat{L}_z \psi\,d\tau = \int D^2 \exp(-r/a)x \hat{L}_z \exp(-r/a)x\,d\tau\,?##

    Edit: I computed this integral, and since f(r)(z+/-ix) are not e.functions of ##\hat{L}_z##, I get terms that vanish and the integral is zero.
     
    Last edited: Nov 25, 2013
  7. Nov 25, 2013 #6

    vela

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    The problem may be asking you to calculate the expectation value of ##\hat{L}_z## as a function of time. It's not clear from the wording, though, so you might want to ask your professor.
     
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