# Variational Method for A Symmetric double well Potential

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.

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 $$$$\psi_0(x) = \frac{1}{\pi^{1/4}}e^{-x^2/2}$$$$ and $$$$H\psi_0(x+a) = \left\{ \frac{1}{2}+\Theta(x)\left[\frac{(x-a)^2}{2}-\frac{(x+a)^2}{2} \right] \right \}\psi_0(x+a)= \left [ \frac{1}{2}- xa \Theta(x) \right ]\psi_0(x+a)$$$$ where $$\Theta(x)$$ is the Heaviside step function which is zero for negative arguments and 1 for positive arguments. $$B_0$$ is $$$$B_0 = \int_{-\infty}^\infty dx \psi_0(x-a)H\psi_0(x+a) = \int_{-\infty}^\infty dx \psi_0(x+a)H\psi_0(x-a)$$$$ with $$\psi_0(x+a)\psi_0(x-a) = \frac{1}{\pi^{1/2}}e^{-x^2-a^2}$$, $$$$B_0 = \frac{e^{-a^2}}{\pi^{1/2}} \left [ \frac{1}{2} \int_{-\infty}^\infty dx e^{-x^2} -2a \int_0^\infty dx xe^{-x^2}\right ] = e^{-a^2}\left [ \frac{1}{2} - \frac{a}{\sqrt{\pi}}\right ]$$$$ which is your required result.
• 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. 2 days ago