Schrodingers Equation and Free Particles

We can show that the wavefunction for a free particle satisfies the Schrodinger wave equation. The process is nowhere near as daunting as the task of 'solving the Schrodinger wave-equation' sounds. In essence, we just have to perform two differentiation operations, and identify the eigenvalue.

For a recap on Schrodinger wave-equation, see here.

`ihbar(del)/(delt)Psi(x,t)=-hbar^2/(2m)(del^2)/(delx^2)Psi(x,t)`

Using the de Broglie wavefunction, described here,

`Psi(x,t)=Aexp(i(kx-omegat+phi))`

In order for our trial wave-function to satisfy the Schrodinger equation, we must show that the wavefunction is an eigenfunction and its energy is an eigenvalue. This means that once we have performed the two differentiations, we obtain back the original wavefunction, multiplied by some new constant. Thus, we start by inserting the definition of our wavefunction, `Psi` into the wave-equation.

`ihbar(del)/(delt)Psi(x,t)=-hbar^2/(2m)(del^2)/(delx^2)(Aexp(i(kx-omegat+phi)))`

1. The right hand side, `delx` derivatives

We expand out the brackets, and noting that A is a constant, we move it outside the differentiation,

`ihbar(del)/(delt)Psi(x,t)=-hbar^2/(2m)A(del)/(delx)(del)/(delx)(exp(i(kx-omegat+phi)))`

To perform the differentation, we use u-substitution. Let `u=i(kx-omegat+phi)`, such that,

`(delu)/(delx)=i((delkx)/(delx)-(delomegat)/(delx)+(delphi)/(delx))`

This is partial differentiation, and we note that the `omegat` does not depend on x, and also that `phi` is a constant, hence,

`(delu)/(delx)=ik` and so, `delx=(delu)/(ik)`

So the right hand side of the Schrodinger equation becomes,

`=-hbar^2/(2m)(ik)A(del)/(delu)(ik) (del)/(delu)exp(u)`

We move the 1/ik constants outside the differentation,

`=-hbar^2/(2m)(ik)^2A(del)/(delu)(del)/(delu)exp(u)`

Recalling that the derivative, `d/dxexp(x)=exp(x)`, then

`=-hbar^2/(2m)(ik)^2Aexp(u)`

We reverse the substitution for u, and arrive at,

`=-hbar^2/(2m)(ik)^2Aexp(i(kx-omegat+phi))`

Immediately obvious is the definition of the initial wavefunction, `Psi(x,t)`, hence we confirm that the wavefunction is indeed an eigenfunction. We now aim to simplify the other terms. Expanding the brackets, squaring i, and canceling the minus signs we obtain,

`=(hbark)^2/(2m)Psi(x,t)`

2. The lhs, `delt` derivative

We start with the left hand side, and insert our wavefunction

`ihbar(del)/(delt)Psi(x,t)=ihbar(del)/(delt)Aexp(i(kx-omegat+phi))`

Once again we have to perform a differentiation, again making use of the substitution, `u=i(kx-omegat+phi)`, and so,

`(delu)/(delt)=i((delkt)/(delt)-(delomegat)/(delt)+(delphi)/(delt))`

This time we note that the `kx` does not depend on t, and also that `phi` is a constant, hence,

`(delu)/(delt)=-iomega` and so, `delt=-(delu)/(iomega)`

Hence the lhs of the wave-equation becomes,

`ihbar(del)/(delt)Psi(x,t)=-i^2hbaromega(del)/(delt)Aexp(u)`

We evaluate i² and cancel the minus signs,

`ihbar(del)/(delt)Psi(x,t)=hbaromega(del)/(delt)Aexp(u)`

We evaluate the exponential derivative,

`ihbar(del)/(delt)Psi(x,t)=hbaromegaAexp(u)`

and again we reverse the substitution for u,

`ihbar(del)/(delt)Psi(x,t)=hbaromegaAexp(i(kx-omegat+phi))`

We recognise the definition of our trial wave-function, `Psi(x,t)`,

`ihbar(del)/(delt)Psi(x,t)=hbaromegaPsi(x,t)`

3. Equating the lhs and rhs

First we restate the Schrodinger wave-equation,

`ihbar(del)/(delt)Psi(x,t)=-hbar^2/(2m)(del^2)/(delx^2)Psi(x,t)`

...and now the result of evaluating the derivatives,

`hbaromegaPsi(x,t)=(hbark)^2/(2m)Psi(x,t)`

Whereas in the former equation, `Psi(x,t)` cannot be eliminated, it can be from the latter. Hence, we arrive at an expression for the energy of the wave, the same as that obtained from the definitions of the de Broglie free-particle formulation,

`hbaromega=(hbark)^2/(2m)`

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