Tricky integral in QM variational method

[akehn]

Homework Statement

Find the variational parameters $\beta$, $\mu$ for a particle in one in one dimension whose group-state wave function is given as:

$\varphi$($\beta$,$\mu$)=Asin(βx)exp(-$\mu$$x^{2}$) for x≥0.

The wavefunction is zero for x<0.

Homework Equations

The Hamiltonian is given as:

H=-$\frac{\\hbar^{2}}{2m}$$\frac{d^{2}}{dx^{2}}$+V(x)

Where the potential field is defined as follows: V(x)=+∞ for x<0
and V(x)=$\frac{-f}{(x+a)^{2}}$ for x≥0

The terms f, a are positive constants.

3. The Attempt at a Solution
I am familiar with the general procedure. I know that

E(β,μ)=<T> + <V>

Where E are the eigen energies and T, V are the kinetic and potential energies, respectively.

To minimize the Hamiltonian one takes partial derivatives of E with respect to β and μ, setting each term equal to zero to determine the variational parameters. This is all straightfoward.

My confusion lies in the evaluation of the normalization coefficient A from <$\varphi$|$\varphi$>=1

The closest tabulated integral has "x" in the exponential term, not "x$^{2}$"

Any help on solving this integral would be greatly appreciated.

 

[gabbagabbahey]

Homework Statement

Find the variational parameters $\beta$, $\mu$ for a particle in one in one dimension whose group-state wave function is given as:

$\varphi$($\beta$,$\mu$)=Asin(βx)exp(-$\mu$$x^{2}$) for x≥0.

The wavefunction is zero for x<0.

Homework Equations

The Hamiltonian is given as:

H=-$\frac{\\hbar^{2}}{2m}$$\frac{d^{2}}{dx^{2}}$+V(x)

Where the potential field is defined as follows: V(x)=+∞ for x<0
and V(x)=$\frac{-f}{(x+a)^{2}}$ for x≥0

The terms f, a are positive constants.

3. The Attempt at a Solution
I am familiar with the general procedure. I know that

E(β,μ)=<T> + <V>

Where E are the eigen energies and T, V are the kinetic and potential energies, respectively.

To minimize the Hamiltonian one takes partial derivatives of E with respect to β and μ, setting each term equal to zero to determine the variational parameters. This is all straightfoward.

My confusion lies in the evaluation of the normalization coefficient A from <$\varphi$|$\varphi$>=1

The closest tabulated integral has "x" in the exponential term, not "x$^{2}$"

Any help on solving this integral would be greatly appreciated.

Hi akehn, welcome to PF!

Integrals of the form $\int_0^\infty e^{-x^2}dx$ can be easily computed by first calculating the square of the integral, transforming the problem from one of calculating a one-dimensional integral to one of calculating a two-dimensional integral over a region, since the 2D integral can be done easily by switching to polar coordinates

$$\begin{align} \left( \int_0^{\infty} e^{-x^2}dx \right)^2 &= \left( \int_0^{\infty} e^{-x^2}dx \right) \left( \int_0^\infty e^{-x^2}dx \right) \\ &= \left( \int_0^{\infty} e^{-x^2}dx \right)\left( \int_0^\infty e^{-y^2}dy \right) \\ &= \int_0^{\infty} \int_0^{\infty} e^{-(x^2 + y^2)}dxdy \\ &= \int_0^{\infty} \int_0^{\frac{\pi}{2}} e^{-r^2}r dr d\theta \end{align}$$

You can utilize this technique for your integral, by using Euler's formula to write $\sin^2 (\beta x)$ in terms of complex exponentials, then "completing the square on the exponent" and making a substitution to get your integrand in the correct form.

 

[akehn]

Ah ha! Thank you very much.

 

[gabbagabbahey]

Ah ha! Thank you very much.

You're welcome! Note that I made a small edit to my post to correct typos, the only one of significance being that the upper limit on the polar angle in the last integral should be $\frac{\pi}{2}$ (90°), not $\frac{\pi}{4}$ (45°).

 

[paranormal]

Thank you for your question. Solving tricky integrals is a common challenge in quantum mechanics, and it is important to approach them carefully and systematically. The key to solving this integral is to first recognize that it can be simplified by using the trigonometric identity sin^{2}(x) = (1-cos(2x))/2. This will allow you to rewrite the wavefunction as follows:

\varphi(\beta,\mu)=A\frac{1-cos(2\beta x)}{2}exp(-\mu x^{2})

Next, you can use the fact that the integral of an odd function over a symmetric interval is zero to simplify the integral. In this case, the interval is from 0 to infinity, so the integral of cos(2\beta x) will be zero. This means that the normalization coefficient A can be determined by setting the integral of the remaining terms equal to 1.

I hope this helps in solving your integral and completing your homework. Remember to always approach tricky integrals with patience and a systematic approach. Good luck!