I am given a set of trial wave functions of the form $$ Φ_n^{\pm}(x)=Ψ_{n}(x-α)\pm Ψ_{n}(x+a) $$ Where $Ψ_n$ are the $n$th Harmonic oscillator wavefunctions. in order to approximate the energy levels from this symmetric Double Well Potential

$$ V(x)=\frac{1}{2}mω^{2}(\lvert x \rvert -a)^2 $$ The hints I have are the following : I am to use the Parity of these trial wavefunctions to reduce the number of integrals I have to calculate for the Energy,and I am supposed to break down the Hamiltonian into this form $$\hat{H_{\pm}}=-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} +\frac{1}{2} mω^2(x \mp α)^2$$ Where the + part holds for $x>0$ and the - part for $x<0$

$$ \hat{H_{\pm}}Ψ_{n}(x \mp a)=\frac{\hbar ω}{2} Ψ_{n}(x \mp a) $$

It is easy to show that the trial wave functions are parity even (+) or odd (-). Using the Parity Operator Left and Right of the Hamiltionian I show that energy values can be simplified from the general form,

$$ \tilde{E_{n}}=\frac{<Φ_{n}^{\pm}|\hat{H}|Φ_{n}^{\pm}>}{<Φ_{n}^{\pm}|Φ_{n}^{\pm}>} $$

To the form

$$ E_{n}^{\pm}=\frac{A_{n} \pm B_{n}}{1+C_n} $$ Where $A_n,B_n,C_n$ are the following integrals

$$B_{n}=<{Ψ_{n}(x+a)}|\hat{H}|{Ψ_{n}(x-a)}>$$ $$A_{n}=<{Ψ_{n}(x+a)}|\hat{H}|{Ψ_{n}(x+a)}>$$ $$C_{n}=<{Ψ_{n}(x+a)}|{Ψ_{n}(x-a)}>$$

Now, I believe I can easily calculate the $C_n$ Integrals by invoking the orthonormality of the Hermite Polynomials when integrating from $-\infty$ to $+\infty$ and the Integral is the same for all values and pretty easy to compute (a simple Gaussian). However, I cannot calculate $A_n$ or $B_n$ correctly no matter what I try. I was given $B_0$ as a result $$B_0=\frac{\hbar ω}{2} (1-2\sqrt{\frac{b}{π}}a)e^{-ba^2}$$

When I try to calculate $B_0$, I break down the integral from $-\infty$ to 0 and from 0 to $+\infty$ and apply the corresponding Hamiltonian to the right wavefunction. As a result, I simply get a factor of $\hbar ω/2$ and get the same integral. I tried to not use the hint and calculate the integral manually I get second degree derivatives of Hermite Polynomials which reduce their class and give zero on all Integrals and some others which seem nearly impossible to calculate. I am not asking for an answer, but I would really appreciate some tips as to how I should start my calculations. Thanks for your time!

New contributor
Jim Charamis is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.

1 Answer 1


I suspect your calculation of $C_n$ is the problem. In any case, you can readily calculate the $B_0$ directly.

Taking natural units where $\hbar = m = \omega = 1$ to simplify the expressions \begin{equation} \psi_0(x) = \frac{1}{\pi^{1/4}}e^{-x^2/2} \end{equation} and \begin{equation} H\psi_0(x+a) = \left\{ \frac{1}{2}+\Theta(x)\left[\frac{(x-a)^2}{2}-\frac{(x+a)^2}{2} \right] \right \}\psi_0(x+a)= \left [ \frac{1}{2}- xa \Theta(x) \right ]\psi_0(x+a) \end{equation} where $\Theta(x)$ is the Heaviside step function which is zero for negative arguments and 1 for positive arguments. $B_0$ is \begin{equation} B_0 = \int_{-\infty}^\infty dx \psi_0(x-a)H\psi_0(x+a) = \int_{-\infty}^\infty dx \psi_0(x+a)H\psi_0(x-a) \end{equation} with $\psi_0(x+a)\psi_0(x-a) = \frac{1}{\pi^{1/2}}e^{-x^2-a^2}$, \begin{equation} B_0 = \frac{e^{-a^2}}{\pi^{1/2}} \left [ \frac{1}{2} \int_{-\infty}^\infty dx e^{-x^2} -2a \int_0^\infty dx xe^{-x^2}\right ] = e^{-a^2}\left [ \frac{1}{2} - \frac{a}{\sqrt{\pi}}\right ] \end{equation} which is your required result.

  • $\begingroup$ I am sorry but I am having a really hard time understanding how the Hamiltonian you used corresponds to the Potential above. If I understand the use of the Step function, it simply limits the values the potential gets to positive x's but the potential inside the brackets reduces to x*a. I apologize if I am missing something obvious here $\endgroup$ Nov 25 at 23:52
  • $\begingroup$ H for $x<0$ is the Hamiltonian for $\psi_0(x+a)$, so for $x<0$ it gives $E_0\psi_0(x+a)$. For $x>0$, H is the Hamiltonian for $\psi_0(x+a)$ after subtracting the terms in the brackets above to correct the potential. $\endgroup$
    – user200143
    2 days ago
  • $\begingroup$ I got it! Thank you very much!I still could not calculate the n-th terms but I made a lot of progress. $\endgroup$ yesterday

Your Answer

Jim Charamis is a new contributor. Be nice, and check out our Code of Conduct.

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.