Ground state energy from the variational method

In summary: I know I'll need it!In summary, the problem asks for the wave function for a particle in a box with potential walls at x=0 and x=L. The variational method is used to find the ground state energy, and it is found to be a constant function with energy 0.
  • #1
rwooduk
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Homework Statement


Consider a one-dimentional particle in a box with infinite potential walls at x=0 and x=L. Employ the variational method with the trial wave function ΨT(x) = sin(ax+b) and variational parameters a,b>0 to estimate the ground state energy by minimising the expression

[tex]E_{T}= \frac{\left \langle \Psi _{T}(x) |H |\Psi _{T}(x) \right \rangle}{\left \langle \Psi _{T}(x) |\Psi _{T}(x) \right \rangle}[/tex]

Homework Equations

The Attempt at a Solution


I calculate the denominator first

[tex]\left \langle \Psi _{T}(x) |\Psi _{T}(x) \right \rangle = \int_{0}^{L} sin^{2}(ax+b) dx[/tex]

I use the trig function

[tex]sin^{2}A = \frac{1}{2} - cos (2A)[/tex]

then integrate to give

[tex] = \frac{L}{2} - \frac{1}{2a}sin(2aL+2b)+\frac{1}{2a}sin(2b)[/tex]

then the numerator

[tex]\left \langle \Psi _{T}(x) |H |\Psi _{T}(x) \right \rangle = \frac{(-i\hbar)^{2}}{2m}\int_{0}^{L}\Psi ^{*}\frac{d^{2}}{dx^{2}} sin(ax+b)dx[/tex]

[tex]= \frac{\hbar^{2}a^{2}}{2m}\int_{0}^{L}sin^{2}(ax+b)dx[/tex]

which is the same integral as before so gets the same result, therefore we have

[tex]E_{T}= \frac{\hbar^{2}a^{2}}{2m}\frac{\frac{L}{2} - \frac{1}{2a}sin(2aL+2b)+\frac{1}{2a}sin(2b)}{\frac{L}{2} - \frac{1}{2a}sin(2aL+2b)+\frac{1}{2a}sin(2b)}[/tex]

which gives

[tex]E_{T}= \frac{\hbar^{2}a^{2}}{2m}[/tex]

this is the bit I'm stuck on, I need to minimise this. if i differentiate with respect to a and set to zero I get a=0.

if there is a potential term it's fine i can do that, but with this question there isnt, any ideas how to minimise it?

as always thanks for any help.
 
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  • #2
I find it strange that you are expected to use variational method for that problem. The trial function ##\sin(ax+b)## is not constructed to automatically satisfy the boundary condition ##\psi(0)=\psi(L)=0##... Where is this problem taken from?

EDIT: I think the idea of this exercise is to demonstrate that the variational method can not be applied blindly without taking into account the boundary conditions relevant for the problem.
 
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  • #3
hilbert2 said:
I find it strange that you are expected to use variational method for that problem. The trial function ##\sin(ax+b)## is not constructed to automatically satisfy the boundary condition ##\psi(0)=\psi(L)=0##... Where is this problem taken from?

EDIT: I think the idea of this exercise is to demonstrate that the variational method can not be applied blindly without taking into account the boundary conditions relevant for the problem.

It's from a past exam paper I'm working through in preparation for exams in January. Here is the complete question:

BXwlDs3.jpg


So would you say that a minimisation of the trial wave function is not needed for this problem? How would you make the statement that it can't be minimised or that the boundary conditions play a part? As I mentioned before when there's a potential term involved, such as for the hydrogen atom minimisation is possible as there are two terms in the minimising differentiation, however I can't see how it works in this problem.

Thanks for the reply and any further ideas would be welcome
 
  • #4
If I was given this exercise, I would just answer that the problem can't be solved because the trial function is invalid due to boundary conditions. If you apply the variational method here without thinking what you are doing, you get the answer that the minumum energy wavefunction is a constant function ##\psi(x)=C## which has energy 0. However, the constant function does not vanish at ##x=0## and ##x=L##, so it is not an acceptable solution.
 
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  • #5
This has had me confused for some time, thanks again for your help!
 
  • #6
If you get the boundary conditions right, on the other hand, you fix b and a just gets a set of discrete set of values, where the smallest one gives exactly the ground state wave function - no variations necessary.
That's a weird problem.
 
  • #7
mfb said:
If you get the boundary conditions right, on the other hand, you fix b and a just gets a set of discrete set of values, where the smallest one gives exactly the ground state wave function - no variations necessary.
That's a weird problem.

The guy running the course is great and his lectures are awesome but he likes to set "interesting" questions on the exam, it really doesn't help when all you want is questions you have done before and recognise. Part (c) of that question is another example of "why does he do this?", just give me the wave fuction and i can do the problem.. I'm not looking forward to the exam, at all.

Thanks again
 

1. What is the variational method?

The variational method is a mathematical technique used to approximate the ground state energy of a quantum mechanical system. It involves choosing a trial function, which is a mathematical representation of the system, and minimizing its energy using variational principles.

2. How does the variational method calculate the ground state energy?

The variational method calculates the ground state energy by finding the minimum energy of the trial function. This is done by solving the variational equations, which are derived from the variational principle.

3. What is the advantage of using the variational method?

The variational method allows for a better approximation of the ground state energy compared to other methods, such as perturbation theory. It also provides a systematic way of improving the energy approximation by using more accurate trial functions.

4. Can the variational method be applied to any quantum mechanical system?

Yes, the variational method can be applied to any quantum mechanical system, including both bound and unbound systems. However, the accuracy of the results may vary depending on the complexity of the system and the choice of trial function.

5. Are there any limitations to the variational method?

One limitation of the variational method is that it can only provide an upper bound for the ground state energy. This means that the actual ground state energy may be lower than the calculated value. Additionally, the accuracy of the results depends on the choice of trial function, so it may not always give the exact ground state energy.

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