Find the normalisation constant using a trial wavefunction

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martinhiggs
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Homework Statement



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

Homework Equations



Choose as a trial wavefunction

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

Verify that

N[tex]^{2}[/tex] = [tex]\frac{K}{16 - 11\alpha + 2\alpha^{2}}[/tex]

The Attempt at a Solution



1 = <[tex]\Psi[/tex]|[tex]\Psi[/tex]>

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

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

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

Homework Statement



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

Homework Equations



Choose as a trial wavefunction

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

Verify that

[tex]N^{2} = \frac{K}{16 - 11\alpha + 2\alpha^{2}}[/tex]

The Attempt at a Solution



[tex]1 = \langle\Psi|\Psi\rangle[/tex]

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

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

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.
 
When I expand the brackets and multiply by x^2

I get the following:

x[tex]^{6} + x^{2} + 2 \alpha^{2} - \alpha^{4} - 2 \alpha x^{3} + \alpha^{2}x^{4} - \alpha x^{5}[/tex]


This seems totally wrong when I look at what I am supposed to get for N^2...
 
martinhiggs said:
When I expand the brackets and multiply by x^2

I get the following:

[tex]x^{6} + x^{2} + 2 \alpha^{2} - \alpha^{4} - 2 \alpha x^{3} + \alpha^{2}x^{4} - \alpha x^{5}[/tex]


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

[tex] \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[/tex]

When you integrate this over the range 0 to 1, you should get the answer.
 
If it helps see a little bit more clearly, [tex]K=210[/tex]. Not sure why your question put [tex]K[/tex] rather than [tex]210[/tex], 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.
 
Ah, excellent! Thank you SO much for your help! :)