QM Wavefunctions: Coefficient Expansion & Linear Combinations

[droedujay]

I found out the Coefficient expansion theorem and constructed the following wavefunction:

Ψ(x,0) = 1/sqrt(2)*Φ1 + sqrt(2/5)*Φ3 + 1/sqrt(10)*Φ5
where φn = sqrt(2/a)*sin(n*pi*x/a)

Is this unique why or why not? I'm thinking that it has something to do with all odd Energies.

Also is

Φ(x) = sin^2 (7*pi*x/a)

an acceptable state. To normalize this function would mean to integrate a sine with a power of 4, how can I do this. Also how can I express this state as a linear combination (mixed states) of energy eigenstates with appropriate coefficients?

 

[Dick]

Your eigenstate wavefunctions are orthonormal. How might this help you in answering the uniqueness question? What would make your candidate wavefunction an acceptible state or not? How might orthonormality help you decompose it into a linear combination of eigenstates? Have you heard of an 'inner product'?

 

[droedujay]

I'm not sure how orthonormality helps with uniqueness.

For second part I know that the candidate wavefunction must fit within the boundaries of the infinite well. This wavefunction:

Φ(x) = sin^2 (7*pi*x/a)

is the initial state, but I don't know where to go from here. The inner product is the linear combination of two vectors

 

[Dick]

The 'inner product' is not a linear combination. <phi_n,phi_m> is the integral of the conjugate of phi_n times phi_m over the interval. Since they are orthonormal it's equal to 1 if n=m and 0 if not. Suppose there were two different expansions of psi(x,0) in terms of the phis and they differed in the value of the coefficient of phi_k. Take the inner product of each with phi_k. Draw a conclusion. Orthonormality implies linear independence.

 

[droedujay]

In order to normalize this wavefunction I would have to expand this wavefunction into an inner product of different wavefunctions. Let's say that phi(1) = sin (7*pi*x/a) and phi(2) = sin (7*pi*x/a). Then the inner product would result in either 1 or 0, but these wavefunctions would have to be different to get a 1 and same to get 0. I'm thinking that coefficients would be different to make them different. Does sin^4 have a linear combination within, for example

(1-cos^2)^2 which could also be written 1 - 2cos^2 + cos^4

what I am trying to figure out is how to expand

Φ(x) = sin^2 (7*pi*x/a)

and would I need to find the normalization constant first? I want to expand this because I can't integrate sin^4.

 

[merryjman]

If sin^4 is indeed the function you have to integrate, then there are lots of integration tables out there that will have the solution you need. I remember that when I took Quantum I spent a lot of time poring over integration tables.

 

[Count Iblis]

Also is

Φ(x) = sin^2 (7*pi*x/a)

an acceptable state. To normalize this function would mean to integrate a sine with a power of 4, how can I do this. Also how can I express this state as a linear combination (mixed states) of energy eigenstates with appropriate coefficients?

Well, it satisfies the boundary conditions and it can be normalized. That's all you need.

I disagree with Merryjman about using integration tables for such simple integrals. How do you think professional physicists and mathematicians have learned to compute integrals? Not by using integration tables! You learn this by doing the integrals yourself as you encouter them in problems.

In this case you can exploit the fact that you are integrating over a multiple of the period of the function. You write

sin(x) = [exp(ix) - exp(-ix)]/(2i)

Take the fourth power of both sides, expand the

[exp(ix) - exp(-ix)]^4

and observe that all terms are zero when integrated from zero to pi, except the constant term (which is binomial(4,2) = 6).

 

[merryjman]

I forgot about Euler's theorem. That would of course be useful here. I also forgot that he was dealing with a finite integral, which would also make it easier. Point conceded :)

 

[Count Iblis]

I forgot about Euler's theorem. That would of course be useful here. I also forgot that he was dealing with a finite integral, which would also make it easier. Point conceded :)

I often encounter third year students who have difficulties doing simple integrations, series expansions etc. They have gotten used to using comper algebra like Mathematica to do the calculus for them from first year onward.

When I see an integral I can usually see within a few seconds how you would go about computing it. This is because I liked to solve difficult integrals since I was 14 years old. So, I have a lot of practice.

When I was studying physics I followed a math course on complex function theory. I was the only physics student attending that course. I scored 95% for the exam. One student scored 70% and all others had failed (they scored less than 60%). They were all complaining how difficult the exam was. But it was just simple contour integration questions. The only difficulty was that you had to figure out which function you would need to integrate over which contour to get the desired result.

I had taught mself countour integration when I was 16 and I had a lot of experience doing that. The (third year) math students had never heard of contour integration before the course. They had to absorb everything from the Cauchy Riemann equations to the Residue Theorem. They were able to understand the proof of the relevant theorems, but at the exam you usually get more practical questions. They simply hadn't had the time to get enough exercises done by exam time...