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Quantum Question

  1. Mar 10, 2009 #1
    A particle of mass m is in a one dimensional well of potential [itex]V(x)=\infty, |x| \geq a[/itex] and [itex]V(x)=0, |x| <a[/itex]

    (i) show the allowed energy eigenvalues are given by

    [itex]E_n = \frac{\hbar \pi^2 n^2}{8ma^2}, n=1,2,...[/itex]

    and determine the normalised eigenfunction [itex]u_n(x)[/itex]
    then sketch the ground state wavefunction.

    DONE THIS BIT

    (ii)Suppose at t=0, the particle has a wavefunction described by

    [itex]\psi(x)=0, |x|>a[/itex] and [itex]\psi(x)=N(a^2-x^2), |x| \leq a [/itex]

    calculate the normalisation constant N and sketch the wavefunction

    (iii)calculate the expectation value of the energy of the particle in state [itex]\psi[/itex] in (ii)

    (iv)for the wavefunciton [itex]\psi[/itex] in (ii), what is the probability that a measurement of energy will yield a value corresponding tot eh ground state.
    interpret your answer.


    So i've done (i) i need someone to leas me through the rest of the questions please.
     
  2. jcsd
  3. Mar 10, 2009 #2
    ii: A wavefunction is normalized when [itex] $\int \psi (x) ^* \psi (x) dx $ [/itex] = 1, so to calculate N, you need to integrate the given wavefunction from negative infinity to infinity (which is effectively the integral from -a to a), equate that to 1, and solve for N.
     
  4. Mar 10, 2009 #3
    iii: In QM, energy is described by an energy operator known as the Hamiltonian. I would start by writing out the Hamiltonian operator. The next step is to recall the formula for the expectation value of an operator in terms of a given wavefunction.
     
  5. Mar 11, 2009 #4
    ok. yeah i was very tired last night - (ii) was very easy.

    so for (iii),

    [itex]\hat{H}=\hat{T}+\hat{V}[/itex] where [itex]\hat{V}=0[/itex] because the wavefunction is 0 for [itex]|x| \geq a[/itex] and so the potential in this region outside the well won't contribute to the expectation.
    do i just let [itex]\hat{T}=-\frac{\hbar^2}{2m}[/itex] ?

    now what?
     
  6. Mar 11, 2009 #5
    You are correct, V = 0 in the region that we're interested in. The kinetic energy operator is given by

    [itex] T = - \frac{ \hbar^2}{2m}\frac{d^2}{dx^2}[/itex]

    So, if we know what the total energy operator is (it's just the kinetic energy operator since there is no potential) we can find it's expectation value. What is the formula for the expectation value of an operator in a particular state?
     
    Last edited: Mar 11, 2009
  7. Mar 11, 2009 #6
    ok just [itex]\langle \psi | \hat{T} | \psi \rangle[/itex]

    great. and any ideas for the last part?
     
  8. Mar 11, 2009 #7
    Well, you have the wavefunction for the ground state from the first part, correct? The following is a very important idea in QM:

    The allowed wavefunctions are by definition eigenfunctions of the Hamiltonian operator H. There is a theorem that says that eigenfunctions of a hermitian operator from an orthonormal basis (when properly normalized). Since the Hamiltonian is hermitian, we know that the allowed wavefunctions form an orthonormal set.

    This means that we can write any generic wavefunction as a linear combination of the energy eigenfunctions. In other words, we can expand:

    [itex]\psi = \sum_{n} c_{n} \phi_{n}[/itex]

    where the phi's are the energy eigenfunctions and psi is any wavefunction (for example, the wavefunction given above).

    The probability of measuring a particular value of an energy in a state [itex]\psi[/itex] is given by the square of the overlap between the state psi and the energy eigenfunction cooresponding to the energy in question. In other words, the probability is given as [itex]|c_{n}|^2[/itex], where c is the coeficient in the above expansion cooresponding to the energy we want to measure.

    So, you first have to find the overlap between the energy ground state and the given state. Do you know how to calculate that overlap?
     
  9. Mar 12, 2009 #8
    so i can just do

    [itex]c_0=\int u*_0 \psi dx[/itex] where [itex]u_0[/itex] is the ground state eigenfunction and then the probability is simply

    [itex]P(E_0)=|c_0|^2[/itex]
    ???

    thanks for your help. i do have one more question though, concerning Pauli Exclusion Principle - pretty elementary, i was just confusing myself over something and need to clear this up:

    it says "no two identical fermions can occupy the same quantum state" - identical here mean they would both have to be the same particle (i.e. two electrons)?

    cheers
     
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