Choosing the trial wavefunction (variational method)

In summary: Psi _{T}(x) |\Psi _{T}(x) \right \rangle}to get ET?In summary, the problem involves determining the appropriate trial wavefunction for an electron confined in a two dimensional region with infinite potential walls, subject to a given potential. The trial wavefunction must satisfy the boundary conditions and have at least one free parameter for the variation method to be used. The chosen trial wavefunction must then be used to calculate the energy using the separation of variables method. The energy is determined by comparing the calculated energies for the x and y components of the trial wavefunction.
  • #1
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Homework Statement


I'm going to list two questions as they offer the same problem with more choices, hopefully it will help realize the method (?) used

(A)
An electron, confined in the two dimensional region 0<x<L and 0<y<L with infinite potential walls, is subject to the potential:

[tex]V(x,y)=V_{1}x + V_{2}y^{2}[/tex]

Which of the following trial wave functions is better suited for approximating the wave function of the electron:

[tex]\psi_{T} (x,y) = cos(ax+by)[/tex]
[tex]\psi_{T} (x,y) = exp -(ax^{2} + by^{4})[/tex]
[tex]\psi_{T} (x,y) = xy(a-x)(b-y)[/tex]

(B)
Consider the infinite potential well V(x) = ∞ at x = ±L/2 and for -L/2 < x < L/2 it is:

[tex]V(x) = \frac{1}{2} k(x-x_{0})^2[/tex]

Which trial wavefunction should we consider:

[tex]\psi_{T} (x) = x(x-L)exp -\gamma(x-x^{0})^2[/tex]
[tex]\psi_{T} (x) = (x- \frac{L}{2} )(x+ \frac{L}{2} )exp -\gamma(x-x_{0})^2[/tex]
[tex]\psi_{T} (x) = \alpha(x- \frac{L}{2} )(x+ \frac{L}{2} ) + \beta exp -\gamma (x-x_{0})^2[/tex]
[tex]\psi_{T} (x) = (x- \frac{L}{2} )^{\alpha}(x+ \frac{L}{2} )^{\beta} + \beta exp -\gamma (x-x_{0})^2[/tex]

More examples from other questions, where the trial wavefunction was already "chosen"

[tex]if... V(r) = A exp (- \frac{r}{a} ) ----> \psi_{T} = c exp (- \frac{\alpha r}{2a} )[/tex]
[tex]if... V(r)= \frac{V_{0} x}{L} ----> \psi_{T} = x(x-L) [/tex]
[tex]if... V(r)=0 ----> \psi_{T} = sin(ax+b) [/tex]

Homework Equations


Intuition? Boundary conditions? Should it be of the same form as the potential?

The Attempt at a Solution



Sometimes it seems the trial wavefunction is picked at random out of a hat, I really don't see how it is chosen. Is there a general method? I asked the lecturer and he said to look at the boundary conditions for the problem, but how would that help me decide?

I really think this question could help a lot of people as I've searched on the internet and most of the time it's simply defined.

Any ideas would be welcome.
 
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  • #2
Can you state the boundary conditions on the wave function for (A) and (B)?
 
  • #3
TSny said:
Can you state the boundary conditions on the wave function for (A) and (B)?

for Question (A) and (B) its an infinite well so I'd assume the boundary conditions to be ψ(0)=ψ(L)=0 that is zero at the boundaries. but can't see how that would help determine the trial wave function.

many thanks for the reply
 
  • #4
Right. The wavefunction must vanish at each point of the boundary. Note that in part (A) the wavefunction is a function of both x and y: ##\psi(x,y)##.

Decide what values of x and y will correspond to being on the boundary of the potential well. Then check each proposed wavefunction to see if the wavefunction vanishes for those values of x and y.

Something looks a little odd in part (A). The statement of the problems specifies 0 < x < L and 0 < y < L. However, the wavefunctions do not contain the constant L. Instead they contain the two constants a and b. I wonder if the statement of the problem should have specified the region as 0 < x < a and 0 < y < b.
 
  • #5
TSny said:
Right. The wavefunction must vanish at each point of the boundary. Note that in part (A) the wavefunction is a function of both x and y: ##\psi(x,y)##.

Decide what values of x and y will correspond to being on the boundary of the potential well. Then check each proposed wavefunction to see if the wavefunction vanishes for those values of x and y.

Thank you I will try this.

TSny said:
Something looks a little odd in part (A). The statement of the problems specifies 0 < x < L and 0 < y < L. However, the wavefunctions do not contain the constant L. Instead they contain the two constants a and b. I wonder if the statement of the problem should have specified the region as 0 < x < a and 0 < y < b.

Ive double checked the question and it's correct, later in the question it does say a and b are free parameters. Here's a link to the paper for your own interest:

https://www.leeds.ac.uk/students/office/exampapers/webserv3_exampapers/phas/13PHYS338101s1.pdf [Broken]

It's question B2 (c)

Thanks for your help!
 
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  • #6
My understanding of the variation method is to choose a trial wavefunction that satisfies the boundary conditions and also has at least one free parameter to vary. In part (A), it appears to me that only one of the proposed functions can satisfy the boundary conditions once a and b are chosen appropriately. But then there will be no free parameter for the variation procedure. Hope I'm not overlooking something. The (B) part looks OK. I think there is one and only one valid choice here.
 
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  • #7
Ok, I'm going to try complete problem (A) (including deriving ET) for anyone else dealing with something similar. I'm just stuck on how to combine the answers at the end. Here's what I've got:

Using the given boundary conditions:

[tex]\Psi _{T}(0,y)=\Psi _{T}(L,y)=0[/tex]
[tex]\Psi _{T}(x,0)=\Psi _{T}(x,L)=0[/tex]

For:

[tex]\psi_{T} (x,y) = cos(ax+by)[/tex]

Applying the boundary conditions gives a=0 and b=0 so this is not appropriate.

For:

[tex]\psi_{T} (x,y) = exp -(ax^{2} + by^{4})[/tex]


Applying the boundary conditions gives a=0 and b=0 so this is not appropriate.

For:

[tex]\psi_{T} (x,y) = xy(a-x)(b-y)[/tex]


Applying the boundary conditions when x=0 Ψ=0 and when y=0 Ψ=0, therefore this is the trial wave function we must use.

Now we define the equation for determining the energy

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

The question states as a hint to use separation of variables so we can write

[tex]\psi_{T} (x,y) = xy(a-x)(b-y)[/tex]

as

[tex]X_{T}(x)= x(a-x)[/tex]
[tex]Y_{T}(y)= y(b-y)[/tex]

Since the equations are similar we can calculate ET for XT(x) and then by comparison determine ET for YT(y), NB the potential term for y is different! These are my results:

[tex]\left \langle X_{T} | X_{T} \right \rangle =\frac{L^{5}}{5}-\frac{a^{2}L^{3}}{3}[/tex]
[tex]\left \langle Y_{T} | Y_{T} \right \rangle =\frac{L^{5}}{5}-\frac{b^{2}L^{3}}{3}[/tex]
[tex]\left \langle X_{T} |T| X_{T} \right \rangle =\frac{\hbar}{m}(\frac{aL^{2}}{2}-\frac{L^{3}}{3})[/tex]
[tex]\left \langle Y_{T} |T| Y_{T} \right \rangle =\frac{\hbar}{m}(\frac{bL^{2}}{2}-\frac{L^{3}}{3})[/tex]
[tex]\left \langle X_{T} |V_{x}| X_{T} \right \rangle = (\frac{L^{6}}{6}-\frac{a^{2}L^{4}}{4})V_{1}[/tex]
[tex]\left \langle Y_{T} |V_{y}| Y_{T} \right \rangle = (\frac{L^{7}}{7}-\frac{b^{2}L^{5}}{5}-\frac{2bL^{6}}{6})V_{2}[/tex]

Question how do I insert these terms in the original equation:

[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}= \frac{\left \langle \Psi _{T}(x) |T |\Psi _{T}(x) \right \rangle + \left \langle \Psi _{T}(x) |V |\Psi _{T}(x) \right \rangle}{\left \langle \Psi _{T}(x) |\Psi _{T}(x) \right \rangle}[/tex]
 
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1. What is the variational method in quantum mechanics?

The variational method is a technique used in quantum mechanics to approximate the ground state energy of a system. It involves choosing a trial wavefunction and minimizing its energy with respect to some parameters. This allows for an upper bound on the true ground state energy to be obtained.

2. Why is choosing the right trial wavefunction important in the variational method?

The trial wavefunction is a crucial component of the variational method, as it determines the accuracy of the energy approximation. Choosing a wavefunction that closely resembles the true ground state wavefunction can lead to a more accurate energy estimation.

3. What factors should be considered when selecting a trial wavefunction?

When selecting a trial wavefunction, factors such as the system's symmetries, boundary conditions, and physical properties should be taken into account. The wavefunction should also be flexible enough to be adjusted and optimized for better results.

4. How can one optimize the trial wavefunction in the variational method?

The trial wavefunction can be optimized by varying its parameters and minimizing its energy. This can be done using techniques such as the gradient descent method or the variational Monte Carlo method. Additionally, using a more complex wavefunction with a larger number of variational parameters can also improve the accuracy of the energy estimation.

5. Is there a guarantee that the variational method will provide an accurate energy estimation?

No, there is no guarantee that the variational method will provide an accurate energy estimation. The accuracy of the method depends on the chosen trial wavefunction and the optimization techniques used. However, as more parameters are added to the trial wavefunction and more optimization iterations are performed, the energy estimation should approach the true ground state energy of the system.

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