Ground state energy from the variational method

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SUMMARY

The discussion focuses on using the variational method to estimate the ground state energy of a one-dimensional particle in a box with infinite potential walls, utilizing the trial wave function ΨT(x) = sin(ax+b). Participants highlight the necessity of satisfying boundary conditions, specifically that Ψ(0) = Ψ(L) = 0, which the chosen trial function does not fulfill. Consequently, the variational method cannot be applied effectively without addressing these conditions, leading to incorrect conclusions about energy minimization. The conversation emphasizes the importance of selecting appropriate trial functions in quantum mechanics problems.

PREREQUISITES
  • Understanding of the variational method in quantum mechanics
  • Familiarity with boundary conditions in wave functions
  • Knowledge of trial wave functions and their properties
  • Basic integration techniques involving trigonometric functions
NEXT STEPS
  • Study the implications of boundary conditions on wave functions in quantum mechanics
  • Learn about valid trial wave functions for different quantum systems
  • Explore the variational principle and its applications in quantum mechanics
  • Investigate the ground state energy calculations for systems with potential terms, such as the hydrogen atom
USEFUL FOR

Students preparing for quantum mechanics exams, particularly those focusing on the variational method and boundary conditions in wave functions. This discussion is beneficial for anyone looking to deepen their understanding of energy minimization techniques in quantum systems.

rwooduk
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Homework Statement


Consider a one-dimensional 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

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

Homework Equations

The Attempt at a Solution


I calculate the denominator first

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

I use the trig function

sin^{2}A = \frac{1}{2} - cos (2A)

then integrate to give

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

then the numerator

\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

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

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

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)}

which gives

E_{T}= \frac{\hbar^{2}a^{2}}{2m}

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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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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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
 
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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This has had me confused for some time, thanks again for your help!
 
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.
 
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
 

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