Find the normalisation constant using a trial wavefunction

In summary, a particle of mass m is confined to a one dimensional potential V(x) between 0 and 1. The particle's wavefunction is given by \Psi(x) = Nx[1 - \alphax + (\alpha - 1)x^{2}]. When asked to verify that this wavefunction satisfies the equation \Psi(x) = Nx[1 - \alphax + (\alpha - 1)x^{2}], the student was able to verify that it does. However, when asked to integrate this wavefunction over the range 0 to 1, they were not able to get the answer correct. They were able to get the answer correct when they multiplied the equation by x^2
  • #1
martinhiggs
24
0

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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  • #2
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.
 
  • #3
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...
 
  • #4
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.
 
  • #5
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.
 
  • #6
Ah, excellent! Thank you SO much for your help! :)
 

1. What is a trial wavefunction?

A trial wavefunction is a mathematical function used to approximate the true wavefunction of a quantum system. It is usually chosen to have a simple form that can be easily manipulated and solved.

2. Why is it necessary to find the normalisation constant?

The normalisation constant ensures that the trial wavefunction satisfies the normalization condition, which states that the integral of the square of the wavefunction over all space must equal 1. This is necessary for the wavefunction to accurately represent the probability distribution of a quantum system.

3. How do you find the normalisation constant?

The normalisation constant can be found by solving the integral of the square of the trial wavefunction over all space and setting it equal to 1. This involves using techniques from calculus, such as integration by parts, to solve the integral.

4. What happens if the normalisation constant cannot be found?

If the normalisation constant cannot be found, it means that the trial wavefunction is not a valid representation of the quantum system. This could be due to errors in the initial assumptions or incorrect manipulation of the wavefunction.

5. Can the normalisation constant be found for any type of trial wavefunction?

Yes, the normalisation constant can be found for any trial wavefunction, as long as it satisfies the normalization condition. However, for more complex wavefunctions, the integration may be more challenging and require advanced mathematical techniques.

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