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Initial conditions in quantum mechanics

  1. Sep 13, 2013 #1
    Hello,

    I'm doing an undergraduate introductory course in quantum mechanics, and I'm having a hard time understanding the interpretation. I've recently learned about the famous "Particle in a box" problem, in which we solve the time-independent schrodinger wave equation to get the energy eigenvalues.

    After solving we find that the particle can only have definite and discrete energy levels. (via an expression in terms of the quantum number n). Now my question is, how do we know which energy level the particle will occupy? Does it depend on initial conditions, if yes then how? or is it that it exists in a superposition of the various states, if yes then how do we calculate the probability factors of each state?

    I have understood how the general form of the wave function for a particular situation describes a given system, but am not able to understand how to incorporate the initial conditions of the system.

    For example, in classical mechanics, we solve the differential equation of motion, which gives us the general form of all solutions, and then we substitute the initial conditions (like initial position, initial velocity etc.) to get the right solution. In the same way how do we do this in QM. How do we take into consideration things like initial position and velocity of the particle. How do we interpret initial conditions.

    I would love to gain a further insight into this concept.
     
  2. jcsd
  3. Sep 13, 2013 #2
    You are probably solving the free particle wave function, that is the Schroedinger equation without any potential energy (except for the walls which are modeled by some boundary condition). Without interactions your particle has no way to jump between states so you have to stipulate by hand which state it is in as an initial condition and the state evolves after that according with the Schroedinger equation.
     
  4. Sep 13, 2013 #3

    vanhees71

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    You are on the right track. The Schrödinger equation (setting [itex]\hbar=1[/itex] for convenience) reads
    [tex]\mathrm{i} \partial_t \psi(t,x)=\hat{H} \psi(t,x).[/tex]
    Now supposed you solved the eigen-value problem for [itex]\hat{H}[/itex] (i.e., the time-independent Schrödinger Equation)
    [tex]\hat{H} \phi_E(x)=E \phi(x).[/tex]
    Since [itex]\hat{H}[/itex] is a self-adjoint operator the eigenvalues are real. We further assume that there are only discrete eigenvalues (the case where the Hamiltonian has also continuous "eigenvalues" or even only continuous "eigenvalues" is only slightly more complicated, if treated in the usual physicist's sloppy way to do mathematics ;-))). Then we have eigenvalues [itex]E_{n}[/itex] and the corresponding eigensolutions [itex]\phi_{n}(x)[/itex]. The eigenfunctions for different eigenvalues are automatically orthogonal to each other, again because of the self-adjointness of [itex]\hat{H}[/itex], and in the case that one or more eigenvalues are equal, we can choose a set of orthogonal vectors spanning the corresponding suspace of eigenvectors for this "degenerate" eigenvalue(s). Further we can assume that the eigenfunctions are normalized, i.e., we have
    [tex]\int \mathrm{d} x \phi_n^*(x) \phi_{m}(x)=\delta_{nm}.[/tex]
    What's not so easy to prove is that for physically reasonable problems these set of solutions are also complete in the sense that we can expand any square-integrable wave function in a series of the energy eigenfunction. So let's assume that.

    Thus we can write the solution of the time-dependent Schrödinger equation as a series expansion wrt. to the energy eigenfunctions,
    [tex]\psi(t,x)=\sum_{n=1}^{\infty} c_n(t) \phi_n(x).[/tex]
    Plugging this into the Schrödinger equation we get
    [tex]\mathrm{i} \sum_{n=1}^{\infty} \dot{c}_n(t) \phi_n(x)=\sum_{n=1}^{\infty} E_n c_n(t) \phi_n(x),[/tex]
    where on the right-hand side we have used the eigenfunction property
    [tex]\hat{H}\phi_n(x)=E_n \phi_n(x).[/tex]
    Now since the [itex]\phi_n[/itex] are orthonormal to each other, the above equation can only be true if the coefficients fulfull the ordinary equations
    [tex]\mathrm{i} \dot{c}_n(t)=E_n c_n(t).[/tex]
    These equations are very simply solved by
    [tex]c_n(t)=c_n(0) \exp(-\mathrm{i} E_n t).[/tex]
    As you see there are infinitely many constants left to be determined. So far our solution of the time-dependent Schrödinger equation reads
    [tex]\psi(t,x)=\sum_{n=1}^{\infty} c_n(0) \exp(-\mathrm{i} E_n t) \phi_n(x).[/tex]
    To uniquely specify the wave function as a solution of the time-dependent Schrödinger equation, we need to know all the coefficients [itex]c_n(0)[/itex].

    Now these coefficients themselves are uniquely defined by the initial condition of the wave function, i.e., we have to specify the wave function at [itex]t=0[/itex]:
    [tex]\psi(t=0,x)=\psi_0(x)=\sum_{n=1}^{\infty} c_n(0) \phi_n(x).[/tex]
    Physically this makes perfect sense, since to know the state of the particle, represented by the wave function, at time [itex]t[/itex] you need to know, in which state this particle was prepared at [itex]t=0[/itex]!

    To find the coefficients we again use the orthonormality of the energy eigenfunctions. Multiply the last equation with [itex]\phi_m^*[/itex] and integrate over [itex]x[/itex], you get
    [tex]\int \mathrm{d}x \phi_m^*(x) \psi_0(x) = \sum_{n=1}^{\infty} c_n(0) \int \mathrm{d} x \phi_m^*(x) \phi_n(x) = \sum_{n=1}^{\infty} c_n \delta_{nm}=c_m(0).[/tex]
    So we have found all the coefficients [itex]c_m(0)[/itex] from the initial wave function, and that solves the Schrödinger equation as an initial-value problem in terms of a "generalized" Fourier series.

    Consider your example of a particle in a potential pot with infinitely large walls. There you had the boundary conditions that the wave function had to vanish at [itex]x=\pm L/2[/itex]. Here the energy eigenvalues are
    [tex]E_n=\frac{n^2 \pi^2}{8 m L^2} \quad \text{with} \quad n \in \mathbb{N}=\{1,\ldots\}.[/tex]
    The eigenfunctions are
    [tex]\phi_n(x)=N_n \times \begin{cases}
    \cos \left (\frac{n \pi x}{L} \right ) & \text{for} \quad n \quad \text{odd},\\
    \sin \left (\frac{n \pi x}{L} \right ) & \text{for} \quad n \quad \text{even}.
    \end{cases}
    [/tex]
    The [itex]N_n[/itex] are normalization factors.

    Here, your series expansion is a usual Fourier series for functions, vanishing at [itex]x=\pm L/2[/itex], i.e., a complete Fourier series with the constant function omitted from the full set of Fourier series. This is so, because of the boundary condition that the wavefunction has to vanisch at [itex]x \pm 1/2[/itex]. It's clear from the general theorems about Fourier series that indeed any sufficiently friendly function with these boundary conditions can be expanded in this way.
     
  5. Sep 13, 2013 #4

    jtbell

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    The initial condition for the time-dependent Schrödinger equation is the wave function at t = 0: ##\Psi(x,0)##.

    As an analogy to the "particle in a box", consider a stretched string whose ends are held fixed, e.g. a violin string. The initial condition for the differential wave equation in this case is the configuration of the string at t = 0, produced e.g. by grabbing the string at some point and pulling it away from equilibrium, stretching it into a shape described by some function y(x,0) whose form depends on exactly where you grab the string. When you let go, the string starts to vibrate in a superposition of its allowed modes (harmonics), with different amplitudes for different modes.

    Similarly with the "particle in a box", you start by "stretching" the wave function into some initial shape, then "let go". The time-dependent wave function is a superposition of the allowed eigenstates, with amplitudes that depend on the precise shape of the initial wave function. The energy of this state is indefinite, with different probabilities of getting the different allowed energies when you make an energy measurement. Those probabilities are determined by the amplitudes of the eigenstates in the superposition.

    For the mathematical details, see vanhees' post. :smile:
     
  6. Sep 14, 2013 #5

    meBigGuy

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    Gold Member

    Great high level view
     
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