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(-\mux^{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(-\mux^{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 &amp;= \left( \int_0^{\infty} e^{-x^2}dx \right) \left( \int_0^\infty e^{-x^2}dx \right) \\ &amp;= \left( \int_0^{\infty} e^{-x^2}dx \right)\left( \int_0^\infty e^{-y^2}dy \right) \\ &amp;= \int_0^{\infty} \int_0^{\infty} e^{-(x^2 + y^2)}dxdy \\ &amp;= \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°).