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Sudden barrier removal to half harmonic oscillator

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data
    A particle is in the ground state of a half harmonic oscillator (V=m/2 w^2 x^2 x>0, and infinity x<0). At t=0, the barrier at x=0 is suddenly removed. Find the possible energy measurements as a function of time and the wavefunction for all times.


    2. Relevant equations
    <H> = <psi|H|psi>, |psi(t)>=Ʃexp(-i*H*t/hbar)|psin(t)> cn


    3. The attempt at a solution
    At t=0, the wavefunction is in the 1st excited state of the full harmonic oscillator, and due to normalization in the half harmonic oscillator, the ground state of the half HO is √2 times the first excited state for the full one. Thus, |psi(0)> = 1/2^0.5 * |psi1>, and so cn=0 for n≠1 and cn=1/√2 for n=1. We then have |psi(t)> = 1/2^0.5 * |psi1> * exp(-i*(3/2 hb w)t/hb). So, |psi(t)> = 1/2^0.5 * |psi1> * exp(-i*(3/2 w)t). This doesn't seem right to me, since the particle is in such a trivial linear combination of states, but I don't see where I have gone wrong, and it makes sense that the expected energy is unchanged. So, can someone point out if there is a mistake? The possible energies to be measured are only 3/2 hbar w, since the particle is in a trivial superposition of just |psi1>

    Thanks
     
  2. jcsd
  3. Oct 11, 2011 #2

    vela

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    The mistake you're making is assuming the system is in the first excited state of the harmonic oscillator. Its wave function isn't [itex]\psi_1(x)[/itex]; it's actually
    [tex]\psi(x) = \left\{\begin{array}{ll}\psi_1(x) & \mathrm{if}~x>0 \\ 0 & \mathrm{if}~x \le 0\end{array}\right.[/tex]at t=0 (up to normalization).
     
  4. Oct 12, 2011 #3
    Thanks, that's true. But still for x>0, psi is just psi1, and for x<0, since the Hermite polynomials are linearly independent, no non-trivial linear combination of them can be equal to zero. So, the energy eigenvalue is still only 3/2 hbar w, then?

    Thanks,
     
  5. Oct 12, 2011 #4

    vela

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    No, that's not correct. You would get only the trivial solution if the linear combination were equal to 0 for all x.
     
  6. Oct 12, 2011 #5
    Ok, so we need to express psi in a basis of the eigenfunctions of the SHO Hamiltonian. for x>0, it is simple as psi1. For x<0, we can't construct a non-trivial linear combination of Hermite polynomials to be zero. So, how can we express psi as nontrivial for x<0.
     
  7. Oct 12, 2011 #6

    vela

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    You can't look at it x<0 and x>0 separately. The same expansion has to hold for all x.
     
  8. Oct 13, 2011 #7
    Thanks again,

    Can we say psi(x) = psi1(x)/T(x), where T(x) = infinity for x<0 and T(x)=1 for x>0. psi(x) is itself still continuous, but then psi'(x) is not, so I guess not.

    But, unless the coefficients themselves are functions of x, I don't see how it would be possible to construct an expansion valid for all x.
     
  9. Oct 13, 2011 #8

    vela

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    Think Fourier series. It's the same idea.

    You have the state [itex]\vert \psi \rangle[/itex], and you want to expand it in terms of the eigenstates [itex]\vert 0 \rangle[/itex], [itex]\vert 1 \rangle[/itex], [itex]\vert 2 \rangle[/itex], etc.
    [tex]\vert \psi \rangle = c_0 \vert 0 \rangle + c_1 \vert 1 \rangle + c_2 \vert 2 \rangle + \cdots[/tex]To solve for cn, you take advantage of the orthonormality of the eigenstates:
    [tex]\langle n \vert \psi \rangle = c_0\langle n \vert 0 \rangle + c_1\langle n \vert 1 \rangle + c_2\langle n \vert 2 \rangle + \cdots = c_n [/tex]To connect this back to the wave functions, insert a complete set to get
    [tex]c_n = \langle n \vert \psi \rangle = \int_{-\infty}^\infty dx\,\langle n \vert x\rangle\langle x \vert \psi \rangle = \int_{-\infty}^\infty \phi_n^*(x)\psi(x)\,dx[/tex]where [itex]\phi_n(x)[/itex] is the eigenfunction for the nth state.
     
  10. Oct 13, 2011 #9
    Right, but the above integral reduces then to Integral from 0 to inf. of phin* psi(x) dx, since psi(x)=0 for x<0 and psi(x) is phi1 for x>0, and so all cn are zero except c1, since the above integral is the inner product of two elements of an L^2 Inner product space, and we are considering inner product of basis elements of that space, which are orthogonal. Is there some mistake?
     
  11. Oct 13, 2011 #10

    vela

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    The orthogonality relation for the Hermite polynomials only holds when you integrate the polynomials from -infinity to infinity. It doesn't hold when you integrate from 0 to infinity. For instance, with H0(x)=1 and H1(x)=2x, you get
    [tex]\int_0^\infty H_0(x)H_1(x)e^{-x^2}\,dx = \int_0^\infty 2xe^{-x^2}\,dx=1[/tex]whereas
    [tex]\int_{-\infty}^\infty H_0(x)H_1(x)e^{-x^2}\,dx = \int_{-\infty}^\infty 2xe^{-x^2}\,dx=0[/tex]
     
  12. Oct 14, 2011 #11
    Of course! Thanks again, I can solve it from here.
     
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