Hamiltonian operator

Homework Statement

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)' ?

The Attempt at a Solution

Related Advanced Physics Homework Help News on Phys.org
What would the eigenvalue equation for H' look like? What if you wrote it in terms of H?

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):
$$\Psi'=e^{\frac{-i\alpha t}{\hbar}}\Psi. (Eq3)$$
$$i\hbar\frac{\partial}{\partial t}\Psi'=e^{\frac{-i\alpha t}{\hbar}}(\hat{H}+C)\Psi.(Eq4)$$
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.

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