Question about when to use the Variational Method

[rwooduk]

Here is the question:

Find the complete spectrum (eigenvalues and eigenstates) of a particle in one dimension described by the Hamiltonian [tex]H = \frac{p^{2}}{2m}+ \frac{1}{2}\gamma (x - a)^{2} + K(x - b)[/tex]
where m,gamma,K,a,b are constants with K>gamma * a and x,p are the positionand momentum of the particle respectively.

We are not given the wavefunction so in this instance would I use the variational method? i.e. should I guess the wavefunction and apply:

[tex]E_{guess} = \frac{< \Psi_{guess} \mid H \mid \Psi_{guess} > }{< \Psi_{guess} \mid \Psi_{guess} >}[/tex]

I am really unsure about how to apply the Hamiltonian, are the x terms position operators since they are a part of the hamiltonian? do you have to expand the squared brackets? the momentum operator (again I am assuming its the operator because its in the Hamiltonian) is squared, so you have to do 2 derivatives of the guessed wavefunction?

Maybe I have misunderstood the question. I'm really confused so any pointers in the right direction would be very much appreciated.

 

[atyy]

Is this a homework problem?

 

[nrqed]

Here is the question:
We are not given the wavefunction so in this instance would I use the variational method? i.e. should I guess the wavefunction and apply:

[tex]E_{guess} = \frac{< \Psi_{guess} \mid H \mid \Psi_{guess} > }{< \Psi_{guess} \mid \Psi_{guess} >}[/tex]

I am really unsure about how to apply the Hamiltonian, are the x terms position operators since they are a part of the hamiltonian? do you have to expand the squared brackets? the momentum operator (again I am assuming its the operator because its in the Hamiltonian) is squared, so you have to do 2 derivatives of the guessed wavefunction?

Maybe I have misunderstood the question. I'm really confused so any pointers in the right direction would be very much appreciated.

They want the exact result, so simply see if you can do a change of variable for x that will bring the hamiltonian in the form of something you already know (plus a constant)

 

[rwooduk]

Is this a homework problem?

Hi, no its not for any assessed homework, the course is 100% exam based. The question was given in a problem set and I'm trying to see how (or if) it relates to the variational method we are covering at the moment.

They want the exact result, so simply see if you can do a change of variable for x that will bring the hamiltonian in the form of something you already know (plus a constant)

Hi, Thanks, I can see that it resembles the Hamiltonian for the simple harmonic oscillator, but isn't x an operator? is it ok to change it?

thanks for the help

 

[nrqed]

Hi, no its not for any assessed homework, the course is 100% exam based. The question was given in a problem set and I'm trying to see how (or if) it relates to the variational method we are covering at the moment.
Hi, Thanks, I can see that it resembles the Hamiltonian for the simple harmonic oscillator, but isn't x an operator? is it ok to change it?

thanks for the help

If we are working in coordinate space, x is just a variable so it is ok to shift it. You are right, this is essentially the harmonic oscillator. The wave functions are the same but the energy is shifted by a constant (which you can work out)

 

[rwooduk]

If we are working in coordinate space, x is just a variable so it is ok to shift it. You are right, this is essentially the harmonic oscillator. The wave functions are the same but the energy is shifted by a constant (which you can work out)

So you're saying since the hamiltonian is that of a simple harmonic oscillator I should use the associated SHO wavefunction for this problem? You don't have to guess what the wavefunction would be? I don't understand why a wavefunction isn't given in the question, does the Hamiltonian determine what the wavefunction should be? (doesnt make sense, seems backwards i.e. shouldn't the wavefunction should determine what Hamiltonian you use) And the question says to find the eigenstates (wavefunctions)

also:

May I ask a side question on correct interpretation of bra and ket?

[tex]< \Psi_{guess} \mid H \mid \Psi_{guess} > = \int \Psi^{*} H \Psi dv [/tex]

but

[tex]\mid H \mid \Psi_{guess} > \neq \int H \Psi dv [/tex]

is that correct? i.e. you need a complete bra ket to interpret it as an integral?

thanks again for any help

 

[rwooduk]

bump, does anyone have any more suggestions?

 

[BvU]

Hamiltonian doesn't determine the wavefunction. All it does is determine how a wavefunction develops in time, namely through the (time dependent) Schroedinger equation.

Separation of variables leads to interest in time-independent solutions of the Schroedinger equation and triggers the hunt for eigenfunctions of the Hamiltonian. For these we write the (time-independent) S.E: Hψ = Eψ.

The wave function does not determine the Hamiltonian. The wave function is a description of a state. What happens to the state depends on the circumstances: one can have kinetic energy conservation, total energy conservation, etc.

Re bra-ket interpretation: You can't write | H |ψ> : the left | is part of the bra. H |ψ> is a state resulting from letting operator H work on state |ψ>.

For eigenfunctions (steady states) the operator is a simple multiplcation.

 

[rwooduk]

Hamiltonian doesn't determine the wavefunction. All it does is determine how a wavefunction develops in time, namely through the (time dependent) Schroedinger equation.

Separation of variables leads to interest in time-independent solutions of the Schroedinger equation and triggers the hunt for eigenfunctions of the Hamiltonian. For these we write the (time-independent) S.E: Hψ = Eψ.

The wave function does not determine the Hamiltonian. The wave function is a description of a state. What happens to the state depends on the circumstances: one can have kinetic energy conservation, total energy conservation, etc.

Re bra-ket interpretation: You can't write | H |ψ> : the left | is part of the bra. H |ψ> is a state resulting from letting operator H work on state |ψ>.

For eigenfunctions (steady states) the operator is a simple multiplcation.

Thats very helpful thanks!

Also I found a useful youtube example for those interested:



THANKS ALL!