Question

prove Schrodinger's equation​

Answer 1

Answer:

Schrodinger's equation cannot be derived. It was thought up using logical arguments and so far it has seemed to work experimentally.

The equations is essentially a re-write up for energy conservation:

E=T+V

Where T is the Kinetic Energy and V is the potential. However, to be more explicit we must work with operators (if you are unsure what operators are I suggest you look them up; this will give you a better understanding of what's going on).

The KE for a particle is given by the KE Operator:

T^=−ℏ22m∂2∂x2

.

This comes from the momentum operator of the particle/wave p^=−ih∂/∂x. You use this in the analogous classical mechanics equation for KE to obtain T^ (Try doing this as an exercise).

So now we are left with just putting it all together. The first equation turns into:

−ℏ22m∂2Ψ∂x2+V(x)Ψ=EΨ

And we define the Hamiltonian H^ as:

H^=−ℏ22m∂2∂x2+V(x)

Thus:

H^Ψ=EΨ

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Answer 2

$\huge\mathfrak\red{Answer}$

Schrodinger's equation cannot be derived. It was thought up using logical arguments and so far it has seemed to work experimentally.

The equations is essentially a re-write up for energy conservation:

E=T+V

Where T is the Kinetic Energy and V is the potential. However, to be more explicit we must work with operators (if you are unsure what operators are I suggest you look them up; this will give you a better understanding of what's going on).

The KE for a particle is given by the KE Operator:

T^=−ℏ22m∂2∂x2

.

This comes from the momentum operator of the particle/wave p^=−ih∂/∂x. You use this in the analogous classical mechanics equation for KE to obtain T^ (Try doing this as an exercise).

So now we are left with just putting it all together. The first equation turns into:

−ℏ22m∂2Ψ∂x2+V(x)Ψ=EΨ

And we define the Hamiltonian H^ as:

H^=−ℏ22m∂2∂x2+V(x)

Thus:

H^Ψ=EΨ

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