Variational Method for A Symmetric double well Potential

Question

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,

$$ \tilde{E_{n}}=\frac{\langle Φ_{n}^{\pm}|\hat{H}|Φ_{n}^{\pm}\rangle}{\langle Φ_{n}^{\pm}|Φ_{n}^{\pm}\rangle} $$

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

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

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!

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.