Find the normalisation constant using a trial wavefunction

[martinhiggs]

Homework Statement

a particle of mass m, confined to a one dimensional infinite potential of
0\leqx\leq1 V(x) = 0
elsewhere V(x) = \infty

Homework Equations

Choose as a trial wavefunction

\Psi(x) = Nx[1 - \alphax + (\alpha - 1)x^{2}]

Verify that

N^{2} = \frac{K}{16 - 11\alpha + 2\alpha^{2}}

The Attempt at a Solution

1 = <\Psi|\Psi>

1 = \int^{1}_{0}Nx[1 - \alphax + (\alpha - 1)x^{2}] Nx[1 - \alphax + (\alpha - 1)x^{2}] dx

1 = N^{2} \int^{1}_{0} x^{2}[1 - \alphax + (\alpha - 1)x^{2}]^{2}

Is this right so far?? I'm not sure how to carry on. Should I expand the brackets??

 

[jdwood983]

Homework Statement

a particle of mass m, confined to a one dimensional infinite potential of
0\leq x \leq 1, V(x)=0
elsewhere V(x)=\infty

Homework Equations

Choose as a trial wavefunction

\Psi(x) = Nx(1 - \alpha x + (\alpha - 1)x^{2})

Verify that

N^{2} = \frac{K}{16 - 11\alpha + 2\alpha^{2}}

The Attempt at a Solution

1 = \langle\Psi|\Psi\rangle

1 = \int^{1}_{0}Nx(1 - \alpha x + (\alpha - 1)x^2)\cdot Nx(1 - \alpha x + (\alpha - 1)x^{2}) \,dx

1 = N^{2}\int^{1}_{0} x^{2}(1 - \alpha x + (\alpha - 1)x^{2})^{2}\,dx

Is this right so far?? I'm not sure how to carry on. Should I expand the brackets??

Yes and Yes. Well, you could possibly integrate it as is, but it'd be easier to see after expanding the parenthesis.

PS: Rather than typing {tex}code{/tex} for every variable, just write the whole thing in tex, it'll look a bit nicer.

 

[martinhiggs]

When I expand the brackets and multiply by x^2

I get the following:

x^{6} + x^{2} + 2 \alpha^{2} - \alpha^{4} - 2 \alpha x^{3} + \alpha^{2}x^{4} - \alpha x^{5}


This seems totally wrong when I look at what I am supposed to get for N^2...

 

[jdwood983]

When I expand the brackets and multiply by x^2

I get the following:

x^{6} + x^{2} + 2 \alpha^{2} - \alpha^{4} - 2 \alpha x^{3} + \alpha^{2}x^{4} - \alpha x^{5}


This seems totally wrong when I look at what I am supposed to get for N^2...

I think you expanded incorrectly. You should get

<br /> \left(\alpha-1\right)^2x^6-2\alpha(\alpha-1)x^5+2(\alpha-1)x^4+\alpha^2x^4-2\alpha x^3+x^2<br />

When you integrate this over the range 0 to 1, you should get the answer.

 

[jdwood983]

If it helps see a little bit more clearly, K=210. Not sure why your question put K rather than 210, but it is what it is.

Also, you could fully expand what I wrote down and then integrate, but you'll have a few extra terms to work with; it's usually easier to expand and regroup after integration.

 

[martinhiggs]

Ah, excellent! Thank you SO much for your help! :)