[Physics] Bound states for sech-squared potential

homework-and-exercisespotentialquantum mechanicsschroedinger equation

I'm working on an introductory qm project, hope somebody has the time to help me (despite the length of this post), it will be highly appreciated.

My goal is to determine the bound states and their energies for the potential

$V_j(x) = -\frac{\hslash^2a^2}{2m}\frac{j(j+1)}{\cosh^2(ax)}$

for any positive value of j (I think I'm supposed to show that j has to be integer at some point, but I don't know). By a change of variable, $ax\rightarrow y$, I have rewritten the Schrödinger equation for this potential to:

$-u''(y)-\frac{j(j+1)}{\cosh^2(y)}u(y) = Eu(y).$

I will denote the corresponding Hamiltonian $H_j$. I have shown that the ground state for $V_j(x)$ is $\psi_0(x) = A\cosh(ax)^{-j}$
(this is all correct, has been verified). Now here is where serious problems begin. I am given these ladder operators:

$a_{j+} = p + ij\tanh(y)$

$a_{j-} = p – ij\tanh(y)$

where $p = -i\frac{d}{dy}$
(no h-bar). I have shown that $H_j = a_{j+}a_{j-} – j^2$
and $H_{j-1} = a_{j-}a_{j+} – j^2$
and I have used this to show that if $|\psi>$
is an eigenstate for $H_j$ then $a_{j-}|\psi>$
is an eigenstate for $H_(j-1)$ and $a_{(j+1)+}|\psi>$
is an eigenstate for $H_(j+1)$, both with the same eigenvalue (energy). Thus the purpose of the a-operators is to carry an eigenfunction for the potential characterized by the value$ j (H_j)$ over to an eigenfunction for the potential characterized by the value j-1 or $j+1 (H_(j-1), H_(j+1)).$

Now I should be able to find all bound states and their energies (I have been told that there is only one bound state for every (integer?) value of j, so this amounts to showing that the ground state I found earlier is the only bound state there is). I'm clueless on this last step.

Best Answer

Assume that you have an arbitrary eigenstate, and apply the lowering operator often enough to be able to say something definite about the resulting vector.

Related Solutions

[Physics] How to predict bound states in a 1D triangular well

Disclaimer: In this answer, we will just derive a rough semiclassical estimate for the threshold between the existence of zero and one bound state. Of course, one should keep in mind that the semiclassical WKB method is not reliable$^1$ for predicting the ground state. We leave it to others to perform a full numerical analysis of the problem using Airy Functions.

     ^ V
    W| 
    a|
    l|
    l|   L
-----|---|------------------> x
     |  /
     | /
     |/
-V_0 | 
     |

$\uparrow$ Fig.1: Potential $V(x)$ as a function of position $x$ in OP's example.

First let us include the metaplectic correction/Maslov index. The turning point at an infinitely hard wall and an inclined potential wall have Maslov index $2$ and $1$, respectively, cf. e.g. this Phys.SE post. In total $3$. We should then adjust the Bohr-Sommerfeld quantization rule with a fraction $\frac{3}{4}$.

$$ \int_{x_-}^{x_+} \! \frac{dx}{\pi} k(x)~\simeq~n+\frac{3}{4},\qquad n~\in~\mathbb{N}_0,\tag{1} $$

where

$$ k(x)~:=~\frac{\sqrt{2m(E-V(x))}}{\hbar}, \qquad V(x)~:=~-V_0 \frac{L-x}{L}. \tag{2} $$

At the threshold, we can assume $n=0$ and $E=0$. The limiting values of the turning points are $x_-=0$ and $x_+=L$. Straightforward algebra yields that the threshold between the existence of zero and one bound state is

$$V_0~\simeq~\frac{81}{128} \frac{\pi^2\hbar^2}{mL^2} \tag{3} .$$

$^1$ For comparison, the WKB approximation for the threshold of the corresponding square well problem yields

$$V_0~\simeq~\frac{\pi^2\hbar^2}{2m L^2} \tag{4} ,$$

while the exact quantum mechanical result is

$$V_0~=~\frac{\pi^2\hbar^2}{8m L^2} \tag{5} ,$$

cf. e.g. Alonso & Finn, Quantum and Statistical Physics, Vol 3, p. 77-78. Not impressive!

     ^ V
    W| 
    a|
    l|
    l|   L 
-----|---|------------------> x
     |   |
     |   |
     |   |
-V_0 |---| 
     |

$\uparrow$ Fig.2: Corresponding square well potential as a function of position $x$. Each of the 2 infinitely hard walls has Maslow index 2.

[Physics] Bound states in semi-infinite potential well

Since you didn't present the complete derivation, I'll just say some general words based on what you showed.

You have infinite values on both sides of the equation, not only on the LHS. Try looking earlier into your derivation. You'll most likely get a sane result if you don't divide by $\cos$ to obtain $\tan$ or if you take a limit of $E\to V_0$ instead of exactly setting $E=V_0$.

If you are just trying to verify that the boundary conditions are satisfied, but can't avoid the division by zero at the boundary for some reason, try setting $E$ early in the process, and compare limits of the wavefunction and of its derivative on both sides to confirm they are equal.

...the energies come out to be quantised by the formula...

This statement sounds very strange: you assert that well depth is quantized. This is absurd. But the equation itself does have a meaning in it, discussed below. I didn't check whether the expression is correct though.

Since $E=V_0$, the wavelength outside of the well is infinite, i.e. the wave function is constant. But the only way you can match a constant with a sinusoid is when the sinusoid is at its extremum. There is only a countable infinity of such sinusoids which are $0$ for $x=L$, and have an extremum at $x=0$. So this means that the state of $E=V_0$ exists only for some well depths (a countable set of such depths), not that the energies are quantized. In fact the energy of this state is not in the discrete part of Hamiltonian's spectrum: the eigenfunction corresponding to this $E$ can't be normalized due to infinitely-extended constant value.

Best Answer

Related Solutions

[Physics] How to predict bound states in a 1D triangular well

[Physics] Bound states in semi-infinite potential well

Related Question