1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Choosing the trial wavefunction (variational method)

  1. Dec 18, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm going to list two questions as they offer the same problem with more choices, hopefully it will help realise the method (?) used

    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]

    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]

    2. Relevant equations
    Intuition? Boundary conditions? Should it be of the same form as the potential?

    3. 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.
  2. jcsd
  3. Dec 18, 2014 #2


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Can you state the boundary conditions on the wave function for (A) and (B)?
  4. Dec 18, 2014 #3
    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 cant see how that would help determine the trial wave function.

    many thanks for the reply
  5. Dec 18, 2014 #4


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    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.
  6. Dec 19, 2014 #5
    Thank you I will try this.

    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!
    Last edited by a moderator: May 7, 2017
  7. Dec 19, 2014 #6


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    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.
  8. Dec 30, 2014 #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]


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

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


    [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.


    [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]


    [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]
    Last edited: Dec 30, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Choosing the trial wavefunction (variational method)
  1. Variation Method (Replies: 3)

  2. Variational method. (Replies: 0)