2
$\begingroup$

I am working through an introductory problem that is part of an undergraduate course for Quantum Theory. I am self-studying currently. $\DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\tanh}{tanh}$

A particle of mass $m$ moving on the $x$-axis has a ground-state wave function $$\psi_1(x)=\sech^2(x).$$ Show that the potential is $$V(x)=-\frac{3\hbar^2}{m}\sech^2(x). $$ If $\psi(x)=\sech(x)\tanh(x)$ is an excited state wave function for the particle, what is the energy of this state?

Both answers seem to follow straightforwardly from the stationary state Schrödinger Equation, and I have obtained $$E=-\frac{\hbar^2}{2m}$$ for the second part (hopefully correct).

From all examples I can find, the energy levels will typically be related by $$E_1n^2=E_n.$$

My question is what is the precise $n$ giving the energy level that has been found? My guess would be the second, since the relation $$\frac{1}{n^2}E_1=E_n$$ could make sense given that the ground state is the inverse of a standard hyperbolic function, and the examples I have covered lead to linear combinations of either $\cos$ & $\sin$, or $\cosh$ & $\sinh$, but I do not feel I am in a position this early on to understand how to explain such a result rigorously.

Please note I have only covered some fundamentals of Quantum Theory, namely:

$\;\;\;\;\;\;\;\;\;\;\;\;\bullet$ 'derivation' of Schrodinger's Equation

$\;\;\;\;\;\;\;\;\;\;\;\;\bullet$ modelling the wave function as a probability density funtion

$\;\;\;\;\;\;\;\;\;\;\;\;\bullet$ simple examples based around 'a particle in a box' (1D and 3D)

$\endgroup$

1 Answer 1

2
$\begingroup$

You were given the information that $\psi_1$ is the wavefunction of a ground state. Ground state, by definition, is the lowest energy stationary state. All stationary states have to fulfill the Time-independent Schrödinger Equation (TISE), so $\psi_1$ also has to. The final piece of the puzzle is that the Hamiltonian of a scalar particle in a potential is $$ H = -\frac{\hbar^2}{2m} \, \frac{\mathrm d^2}{\mathrm dx^2} + V(x) \: , $$ where $V(x)$ is the unknown function. And now it's just algebra: \begin{align} H \, \psi_1 &= E_1 \, \psi_1 \tag{TISE} \\ - \frac{\hbar^2}{2m} \, \frac{\mathrm d^2}{\mathrm dx^2} \, \psi(x) + V(x) \, \psi_1 &= E_1 \, \psi_1 \\[5pt] - \frac{\hbar^2}{2m} \, \frac{\mathrm d^2}{\mathrm dx^2} \, \operatorname{sech}^2(x) + V(x) \, \operatorname{sech}^2(x) &= E_1 \, \operatorname{sech}^2(x) \\[5pt] - \frac{\hbar^2}{2m} \big( 4 \operatorname{sech}^2(x) - 6 \operatorname{sech}^4(x) \big) -E_1 \, \operatorname{sech}^2(x) &= - V(x) \, \operatorname{sech}^2(x) \tag{1} \label{wolfram} \\[5pt] \frac{\hbar^2}{2m} \big( 4 - 6 \operatorname{sech}^2(x) \big) + E_1 &= V(x) \\[5pt] -\frac{3\hbar^2}{m} \operatorname{sech}^2(x) + \Big( E_1 + \frac{2\hbar^2}{m} \Big) &= V(x) \\[5pt] \end{align} At the step marked \eqref{wolfram} I used WolframAlpha (link). Since we weren't told the value of the ground-state energy $E_1$, we can fix it on any value we wish, choosing $E_1 = -2\hbar^2/m$ in particular gives us $$ V(x) = -\frac{3\hbar^2}{m} \operatorname{sech}^2(x) \: . $$ Now we know the entire Hamiltonian and, in principle, we can find all of its energies and stationary states by solving the TISE, this time with $E$ and $\psi$ as unknowns. However, you have it easy, since you're already been told what $\psi$ is, so you just solve for $E$: $$ \frac{\hbar^2}{m} \Bigg( {- \frac{1}{2}} \frac{\mathrm d^2}{\mathrm dx^2} \, -3 \operatorname{sech}^2(x) \Bigg) \, \operatorname{sech}(x) \, \operatorname{tanh}(x) = E \, \operatorname{sech}(x) \, \operatorname{tanh}(x) \: . $$ I trust you can pull this off ;)

Notice that we aren't using the equation $E_1n^2=E_n$. Why? Because it doesn't hold – at least not generally! It is an equation that holds for the energies of a hydrogen atom, but not for a general Hamiltonian. And even the hydrogen atom Hamiltonian has energies $E>0$ for which the equation is useless. To find the energies of a generic system, you just have to find the spectrum of $H$, there is no short way around it.

$\endgroup$
3
  • $\begingroup$ $\DeclareMathOperator{\sech}{sech}$ Thank you for this! I've just realised the mistake is that I accidentally put $\psi(x)=\sech(x)$ when it should have been $\psi(x)=\sech^2(x)$. $\endgroup$
    – 133crem
    Sep 1, 2021 at 17:59
  • $\begingroup$ I have another question up regarding the probability current that I could use some help with if you wouldn't mind taking a look! math.stackexchange.com/questions/4238155/… $\endgroup$
    – 133crem
    Sep 1, 2021 at 18:01
  • $\begingroup$ I've edited the answer to have $\psi(x) = \operatorname{sech}^2(x)$. If this answers your question, please don't forget to mark it as the correct answer :) $\endgroup$
    – m93a
    Sep 1, 2021 at 20:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.