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Hamiltonian operator

  1. Apr 21, 2009 #1
    1. The problem statement, all variables and given/known data
    One dimensional harmonic oscillator has the Hamiltonian
    H(hat)=p(hat)/2m +0.5mw^2x(hat)^2
    Show that the eigenvalue spectrum of H(hat) is
    En=(n+0.5)h(bar)w n=0,1,2...
    I've managed to show this

    Suppose the real constant C is added to the Hamiltonian H(hat) to give the new Hamiltonian
    H(hat)' = H(hat) + C
    What is the eigenvalue spectrum of H(hat)' ?



    2. Relevant equations
    I am afraid I have no idea how to go about this question!


    3. The attempt at a solution
     
  2. jcsd
  3. Apr 21, 2009 #2
    What would the eigenvalue equation for H' look like? What if you wrote it in terms of H?
     
  4. Apr 24, 2009 #3
    Dear captainjack2000, it's really a good problem, and I wish I could do some help.

    Firstly, I'm to re-express the problem. Generally, Hamiltonian operator is connected with the total energy,
    $$
    \hat{H}=\frac{\hat{p}^2}{2m}+V=-\frac{\hbar^2}{2m}\nabla^2+V,
    $$
    hence,
    $$
    \hat{H}'=\frac{\hat{p}^2}{2m}+(V+C)=-\frac{\hbar^2}{2m}\nabla^2+(V+C).
    $$
    So, the problem is indeed equal to:\\

    \textsf{What are the effects on the eigenvalue spectrum of
    Hamiltonian, due to the selection of zero reference level for
    potential energy?}\\

    Now, let's begin the analyses via Schr$\ddot{o}$dinger Equation (SE):
    $$
    i\hbar\frac{\partial}{\partial t}\Psi=\hat{H}\Psi, (Eq1)
    $$
    $$
    i\hbar\frac{\partial}{\partial t}\Psi'=\hat{H}'\Psi'=(\hat{H}+C)\Psi', (Eq2)
    $$
    where $C$ is the difference resulting from the change of reference level for potential energy. Next let's formulate how $\Psi'$ differs from $\Psi$, and we introduce a phase shift factor as an attempt (this is in fact a common method, for example, we turn to this method again when we verify the Galilean invariance of wave equation, or explore the new formalism of wave equation from non-inertial reference under uniform acceleration):
    $$
    \Psi'=e^{\frac{-i\alpha t}{\hbar}}\Psi. (Eq3)
    $$
    Eq2 \& Eq3 lead to
    $$
    i\hbar\frac{\partial}{\partial t}\Psi'=e^{\frac{-i\alpha t}{\hbar}}(\hat{H}+C)\Psi.(Eq4)
    $$
    On the other hand, time derivative of Eq3 gives rise to
    $$
    i\hbar\frac{\partial}{\partial t}\Psi'=e^{\frac{-i\alpha t}{\hbar}}(i\hbar\frac{\partial}{\partial t}+\alpha)\Psi. (Eq5)
    $$
    Comparison of Eq4 and Eq5 results in
    $$
    (\hat{H}+C)\Psi=(i\hbar\frac{\partial}{\partial t}+\alpha)\Psi. (Eq6)
    $$
    Combination of Eq6 \& Eq1 immediately lead to
    $$
    \alpha=C. (Eq7)
    $$
    Hence, Eq3 and Eq4 are rewritten as
    $$
    \Psi'=e^{\frac{-iCt}{\hbar}}\Psi, (Eq8)
    $$
    $$
    i\hbar\frac{\partial}{\partial t}\Psi'=e^{\frac{-iCt}{\hbar}}(\hat{H}+C)\Psi, (Eq9)
    $$
    respectively. Suppose $\Psi$ and $\Psi'$ are for stationary states, then, via separation of variables, we have
    $$
    \Psi(t,r)=e^{-iEt/\hbar}\phi(r) (Eq10)
    $$
    $$
    \Psi'(t,r)=e^{-iE't/\hbar}\phi'(r) (Eq11)
    $$
    where $E$ and $E'$ are real constants, and separately represent the energy of the states $\Psi$ and $\Psi'$. From Eq9, Eq10 and Eq11, we
    have
    $$
    i\hbar\frac{\partial}{\partial t}\Psi' =e^{\frac{-iCt}{\hbar}}(\hat{H}+C)\Psi =e^{\frac{-iCt}{\hbar}}(E+C)\Psi =(E+C)\Psi' (Eq12)
    $$
    where we used
    $$
    \hat{H}\phi(r)=E\phi(r),\sim \hat{H}\phi(r)e^{-iEt/\hbar}=E\phi(r)e^{-iEt/\hbar},\sim \hat{H}\Psi=E\Psi.
    $$
    Hence, to sum up, we attained that
    $$
    \Psi'(t,r)=e^{-iEt/\hbar}\Psi(t,r)
    $$
    $$
    E'=E+C
    $$
    Now, we're ready to get down to the eigenvalue spectrum of the newHamiltonian $\hat{H}'$:
    $$
    \Psi(r,t)=\sum_n C_n\phi_n(r)e^{-iE'_nt/\hbar}.
    $$
    After separation of variables, for $\hat{H}'$, $\hat{H}'=\hat{H}+C$:
    $$
    \left(-\frac{\hbar^2}{2m}\nabla^2+U(r)+C\right)\phi(r)=E\phi(r).
    $$
    For simplicity, we take one-dimensional case as an example:
    $$
    \left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+U(r)+C\right)\phi(x)=E\phi(x).
    $$
    Hence
    $$
    -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\phi+[(E-C)+U(r)]\phi=0.
    $$
    Hence
    $$
    \frac{d^2}{dx^2}\phi+\frac{2m}{\hbar^2}\left[(E-C)+U(r)\right]\phi=0. (Eq13)
    $$
    This is the usual 2-order ODE, from which the spectra of Hamiltonian will be achieved. And before, with the ODE:
    $$
    \frac{d^2}{dx^2}\phi+\frac{2m}{\hbar^2}\left[E+U(r)\right]\phi=0, (Eq14)
    $$
    various special cases (i.e. $U(r)$ is determined in formalism), such as harmonic oscillators, infinite square wells, and we get the corresponding Hamiltonian spectra for each case, which depend on
    $U(r)$:
    $$
    \text{Spectra Set}: {E_n:\quad E=E(n), n=0,1,2\ldots} (Eq15)
    $$
    Comparing Eq13 with Eq14, we just need to replace $E$ with $E-C$ in Eq15, hence,
    $$
    \text{Spectra Set}: {E_n: E-C=E(n), n=0,1,2\ldots},
    $$
    hence,
    $$
    \text{Spectra Set}: {E_n: E=E(n)+C, n=0,1,2\ldots}.
    $$
    Hence, we draw the conclusion that:
    $$
    \hat{H}'=H+C, \sim E'_n=E_n+C
    $$
    This result holds for arbitrary form of $U(r)$, i.e. for arbitrary potential.

    Please translate the standard Latex codes yourself, or contact tianwj1@gmail.com for texified pdf file with visual formula.
     
    Last edited: Apr 24, 2009
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