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I am trying to find energy eigenvalues for a particle in a potential $\ V(x) = Bx^\gamma$ where $\gamma$ is a positive, even integer (2,4,6,8....).

Considering boundary conditions, V(x) will go to infinity as desired due to the even nature of the potential function.

Time independent Schrodinger eqn will give $ (\frac{d^2\Psi}{dx^2}) + (\frac{2mE}{\hbar^2}-\frac{2mB}{\hbar^2}x^\gamma)\Psi = 0$.

Let $\beta = \frac{2mE}{\hbar^2}$

and $\alpha^2 = \frac{2mB}{\hbar^2}$

Now, if we scale x to: $x^\gamma = \frac{u^\gamma}{\alpha^{\gamma/2}}$, we are able to find solutions in terms of $u$.

So that the TISE becomes $ \alpha(\frac{d^2\Psi}{dx^2}) + (\beta-\alpha^2x^\gamma)\Psi= 0$.

If we take the case $\gamma = 2$, we obtain

$ (\frac{d^2\Psi}{du^2}) + (\frac{\beta}{\alpha}-u^2)\Psi= 0$.

Choosing $\frac{\beta}{\alpha} = 2q+1$ and setting it equal to the values of $\beta$ and $\alpha$ we assigned earlier gives the standard result for the simple harmonic oscillator:

$\frac{\beta}{\alpha} = 2q+1 = \frac{2E}{\hbar}\sqrt{\frac{m}{2B}}$

$E_q = (q+\frac{1}{2})\hbar\sqrt{\frac{2B}{m}}$

where frequency $\omega = \sqrt{\frac{2B}{m}}$

Now, I don't really know how to continue with the higher order terms for $\gamma$. Do I need to change my scale factor of $x$ as to isolate $u^4$?

Furthermore, I don't really understand the significance of the higher order potentials. Are they actually relevant, and do they lead to bound state solutions? Is there a single set of energy eigenvalues that is universal to this potential? I don't see how we could obtain a universal set of energy eigenvalues for a potential function that deals with changing orders of magnitude.

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Too long for the comment section but not a complete answer either I'll just post the following insight into the energy eigenvalue quantisations for potential wells of the $V=x^{\gamma}$ type.

Just look at the plots for $\gamma=2$, $\gamma=4$, $\gamma=6$, etc, $\gamma=20$. The higher $\gamma$, the more the potential well resembles the archetypal potential well with infinitely high potential walls. The latter occurs for $\gamma=+\infty$. Below, a plot for $\gamma=40$:

Gamma = 40

For the potential well with infinitely high potential walls, we know that: $$E_n \propto n^2$$ (for $n=1,2,3,...$)

For the quantum harmonic oscillator ($\gamma=2$), we know that: $$E_n\propto n$$ (for $n=0,1,2,3,...$)

It therefore seems reasonable to assume that for:

$$2<\gamma<+\infty$$

The quantisation lies somewhere between $n^1$ and $n^2$. All should have an infinite number of bound states.

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  • $\begingroup$ $E_n = \frac{n^2\pi^2\hbar^2}{2mL}$ for the infinite well, where L is the length between the walls. And $E_n = (n + \frac{1}{2})\hbar\sqrt{\frac{k}{m}}$ for the simple harmonic oscillator. The two solutions for energy states don't feel familiar since they are derived in very different ways. Am I to assume that within $V(x) = x^\gamma$ for $2<\gamma_{even}<∞$ that each value of $\gamma$ will require different, more complex methods than those for the oscillator, and square well energies? Or is there a general solution that describes all of $E$ for the varying potential $x^\gamma$? $\endgroup$
    – bleuofblue
    Commented Oct 7, 2016 at 4:13
  • $\begingroup$ A side question: Comparing the two eigenenergies above for the infinite well and the quadratic oscillator, how would the other constants aside from $n$ change as $\gamma$ increased? For example we notice that in the quadratic oscillator, energy depends on mass as $\frac{1}{\sqrt{m}}$ and for the infinite square well, just $\frac{1}{m}$. Is there anything to suggest that higher order potentials would change other parameters which the energy may depend on? $\endgroup$
    – bleuofblue
    Commented Oct 7, 2016 at 4:14
  • $\begingroup$ Also above from the first comment, the potential $x^\gamma$ essentially becomes the square well at $\gamma = ∞$. So in range of $2<\gamma<∞$, when does this approximation hold? And for an instance like $\gamma = 40$ as you plotted, would the solutions be that of an infinite well with some perturbation? $\endgroup$
    – bleuofblue
    Commented Oct 7, 2016 at 4:24
  • $\begingroup$ There is almost certainly no general algorithm that allows to solve the TISE for ALL values of $\gamma$. My 'insight' only pertains to the shape of the energy spectrum and that in all cases there are $\infty$ bound states. For $\gamma>40$ the square infinite well would be a very good approximation: $E_n\propto n^2$. Thanks. $\endgroup$
    – Gert
    Commented Oct 7, 2016 at 13:35

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