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Energy From Schrodinger Equation

  1. Nov 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Given [itex]\Psi(x, y, z)=(2/L)^{3/2}sin(\frac{n_x\pi x}{L})sin(\frac{n_y\pi y}{L})sin(\frac{n_z\pi z}{L})[/itex], calculate the first few energy levels and tell which are degenerate.

    3. The attempt at a solution

    I don't have much of an attempt to be honest.. What I've done so far is find the Energy Operator [itex]\hat{E}=i\hbar \frac{\partial}{\partial t}[/itex], however, I quickly realized this wouldn't help because my wave function is not time dependent. Next, I realized that this particle has no potential energy (this is an infinite well problem), so [itex]E=\frac{p^2}{2m}[/itex], and [itex]p=\frac{\hbar}{\lambda}[/itex]. However, I also don't know how to calculate lambda and I'm not even sure that I'm headed in the right direction >.<

    I've seen other formulas for energy levels of a particle. Of course, my teacher went through this in class (unfortunately the class has trouble keeping up with him). The problem is, I don't know if those formulas apply to this specific wave function, and furthermore, I should be able to do this from scratch myself, so that's my aim. Any help is appreciated.

    edit-
    From a few things I posted above:
    [itex]E=\frac{\hbar^2}{2m\lambda^2}[/itex]

    Since there are multiple wavelengths, (in each of the three degrees of freedom), can I say:
    [itex]E=\sum{\frac{\hbar^2}{2m\lambda^{2}_{i}}}[/itex]
     
  2. jcsd
  3. Nov 19, 2013 #2

    George Jones

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    Do you know of any other way that p is expressed in quantum theory?
     
  4. Nov 19, 2013 #3
    The only other ways that I know of (or that I can think of) to express p involve the velocity, and I wouldn't know how to calculate the "velocity" of the particle in this example :\

    edit-
    Oh and the momentum operator. Is this what you're referring to?
    [itex]\hat{p}=-i\hbar \Delta^{2}[/itex]

    edit2-
    I realize my symbol is upside down. Not worth the effort right now! Haha
     
  5. Nov 19, 2013 #4

    George Jones

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    In your course, you haven't seen expressed as an operator?
     
  6. Nov 19, 2013 #5
    Yes I do remember that. Interesting. How come the momentum operator applies here but the energy operator doesn't? (I will see if I can get the solution now)

    edit-
    Using the momentum operator, I found the total energy to be:
    [itex]E=C(n_x^2+n_y^2+n_z^2)\Psi^2[/itex]

    But as you can see, this still depends on x, y and z because it still depends on Psi. In the other energy equations I've seen it's a total energy, not just the energy at a point.. Does this mean I need to integrate this energy over the volume of the box? Is there an easier way? o_O

    edit2-
    Blah! I just realized momentum is not a Laplacian, it's a gradient >.<!! I've got some erasing in my near future. The question still stands though.. Is there an easier way? The gradient squared will still return a function of x, y, and z..

    edit3-
    Disregard previous edits. I've been experimenting with operators a lot and I've realized that I don't understand them very well!

    I'm looking at the derivation on wikipedia of the momentum operator in one dimension
    HERE

    My first question is:
    They used the plane wave solution of the Shrodinger equation to derive it. So does it apply to any wave function or just the plane wave? (Plane wave is U = 0 right?)

    Second question:
    I'm looking at:
    [itex]\frac{\partial \Psi}{\partial x} = i \frac{p}{\hbar} \Psi[/itex]
    Solving for p, I get [itex]p=\frac{-i\hbar\frac{\partial \Psi}{\partial x}}{\Psi}[/itex]
    So it's clear from this that the momentum is:
    [itex]p=\frac{\hat{p}\Psi}{\Psi}[/itex]
    Is this true for energy as well? If I want to find energy do I apply the energy operator and then divide by the wavefunction? Is it like this for all operators? What's the general rule?
     
    Last edited: Nov 19, 2013
  7. Nov 19, 2013 #6

    vela

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    Not all states will satisfy ##\hat{p}\psi = p\psi## where ##p## is some constant. Those that do are called momentum eigenstates. They have a definite momentum, namely ##p##, so it makes sense to talk about THE momentum of such a state.

    Say you have ##\psi(x) = e^{ikx}##. If you apply ##\hat{p}## to it, you get
    $$\hat{p}\psi(x) = -i\hbar \frac{\partial}{\partial x} e^{ikx} = (\hbar k) \psi(x).$$ This result tells you that ##\psi(x)## is a momentum eigenstate, and the constant ##\hbar k## multiplying ##\psi(x)## at the end is the momentum associated with that state. On the other hand, if you had the state ##\phi(x) = \sin(kx)##, applying ##\hat{p}## yields
    $$\hat{p}\phi(x) = i\hbar k \cos(kx).$$ Clearly, the righthand side is not a multiple of ##\phi(x)##, so ##\phi(x)## isn't a momentum eigenstate, and the state doesn't have a definite momentum.

    The time-independent Schrodinger equation
    $$\hat{H}\psi(x) = E\psi(x)$$ is an eigenvalue equation with the energy E as the eigenvalue. You want to come up with the appropriate operator ##\hat{H}##, which is called the Hamiltonian, and apply it to the given wave function to verify that it's indeed an energy eigenstate and to identify the eigenvalue E.
     
  8. Nov 19, 2013 #7

    vanhees71

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    Although there is some danger in confusing you even more, I must stress that for the infinite-square-well problem there is no momentum operator defined. This is a pretty subtle point that is often overlooked, and I think it's even wrong in some textbooks.

    The point is that an operator that is representing an observable has to be not only Hermitean but also self-adjoint, i.e., if there was a momentum operator, defined to generate spatial translations, in position representation it would have the form [itex]\hat{p}=-\mathrm{i}/\hbar \vec{\nabla}[/itex]. The space of wave functions is [itex]L^2([0,L]^3)[/itex] with the boundary conditions [itex]\psi(0)=\psi(L)=0[/itex].

    Further, if [itex]\hat{p}[/itex] was self-adjoint it should have a real spectrum, but the wave functions would be
    [tex]u_{\vec{p}}(\vec{x})=N \exp(\mathrm{i} \vec{x} \cdot \vec{p}/\hbar),[/tex]
    but for no value of [itex]\vec{p}[/itex] you can fulfill the boundary conditions, i.e., there are no (generalized) eigenvectors within the Hilbert space of your problem, and thus [itex]\hat{p}[/itex] cannot be a self-adjoint operatator.
    This argument does not hold for the Hamilton operator of this problem, which indeed is
    [tex]\hat{H}=-\hbar^2 \frac{\Delta}{2m}.[/tex]
    This operator is indeed selfadjoint, and a complete set of eigenfunctions is given by the functions in the OP, which fullfills both
    [tex]\hat{H} u_{n_x,n_y,n_z}(\vec{x})=E(n_x,n_y,n_z) u_{n_x,n_y,n_z}(\vec{x}), \quad E(n_x,n_y,n_z) =\frac{\hbar^2 \pi^2}{2m}(n_x^2+n_y^2+n_z^2), \quad (n_x,n_y,n_z) \in \mathbb{N}^3[/tex]
    and the boundary conditions of the infinite-square-well. This means that [itex]\hat{H}[/itex] is a valid operator to represent an observable, and since it's also positive definite, it's a good candidate for a Hamiltonian.
     
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