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Wave Functions help

1) Ψ1 = sin(𝑘𝑥)
2) Ψ2 = e^𝑖𝑘𝑥 = 𝑐𝑜𝑠(𝑘𝑥) + 𝑖sin(𝑘𝑥)

For each wave function, show that they are eigenfuntions of the hamiltonian in
i) Free Space (V(x)=0).
ii) A 'flat' potential (V(x) = V)

In each case, what is the kinetic and total energy?

I've done all of it, but how do I show in each case, what is the kinetic and total energy?

Thanks!
Reply 1
Original post by PencilPot!
1) Ψ1 = sin(𝑘𝑥)
2) Ψ2 = e^𝑖𝑘𝑥 = 𝑐𝑜𝑠(𝑘𝑥) + 𝑖sin(𝑘𝑥)

For each wave function, show that they are eigenfuntions of the hamiltonian in
i) Free Space (V(x)=0).
ii) A 'flat' potential (V(x) = V)

In each case, what is the kinetic and total energy?

I've done all of it, but how do I show in each case, what is the kinetic and total energy?

Thanks!


What's the Hamiltonian operator in each case?
Reply 2
Original post by alow
What's the Hamiltonian operator in each case?


i) ħ^2k^2/2m sin(kx)

ii) sin(kx) [ħ)^2k^2/2m + V]

i) -ħ^2 i^2 k^2/2m e^ikx

ii) e^ikx [-ħ^2 i^2 k^2/2m + V]
(edited 6 years ago)
Reply 3
Original post by PencilPot!
i) ħ^2k^2/2m sin(kx)

ii) sin(kx) [ħ)^2k^2/2m + V]

i) -ħ^2 i^2 k^2/2m e^ikx

ii) e^ikx [-ħ^2 i^2 k^2/2m + V]


Ignoring constants, the Scrodinger equation will be:

H^Ψn=[12d2dx2+V(x)]Ψn=EnΨn\hat{H} \Psi_n = \left[ -\dfrac12 \dfrac{\text{d}^2}{\text{d}x^2} + V(x) \right] \Psi_n = \text{E}_n \Psi_n

12d2dx2- \dfrac12 \dfrac{\text{d}^2}{\text{d}x^2} is the kinetic energy operator part of the Hamiltonian.

So for V=0V=0, how much potential energy is there?
Reply 4
Original post by alow
Ignoring constants, the Scrodinger equation will be:

H^Ψn=[12d2dx2+V(x)]Ψn=EnΨn\hat{H} \Psi_n = \left[ -\dfrac12 \dfrac{\text{d}^2}{\text{d}x^2} + V(x) \right] \Psi_n = \text{E}_n \Psi_n

12d2dx2- \dfrac12 \dfrac{\text{d}^2}{\text{d}x^2} is the kinetic energy operator part of the Hamiltonian.

So for V=0V=0, how much potential energy is there?


1/2?
Reply 5
Original post by PencilPot!
1/2?


V(x) is the potential energy operator so when it is zero the potential energy will be...
Reply 6
Original post by alow
V(x) is the potential energy operator so when it is zero the potential energy will be...


0 :biggrin:
Reply 7
Original post by PencilPot!
0 :biggrin:


Yep.

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