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 im 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)

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 isnt x an operator? is it ok to change it?

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 dont have to guess what the wavefunction would be? I dont understand why a wavefunction isnt given in the question, does the Hamiltonian determine what the wavefunction should be? (doesnt make sense, seems backwards i.e. shouldnt 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?

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.