1. Show that (x) = Asin(kx) satisfies the time-independent Schrödinger equation when E and V are independent of x?

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NJ Share
Mar 9, 2018

You have provided a time-independent wavefunction, where the wavefunction is only a function of some dimension $x$:

$\setminus \Psi \left(x\right) = A \sin \left(k x\right)$

The time independent Schrödinger equation, assuming a single nonrelativistic particle is:

$\setminus \hat{H} \setminus \Psi \left(\setminus \vec{r}\right) = E \setminus \Psi \left(\setminus \vec{r}\right)$

Of course this equation states that the time-independent wavefunction is an eigenfunction of the Hamiltonian corresponding to an eigenvalue equal to the energy.

We can write the TI-equation expanded as:

[(-ħ^2)/(2m)\grad^2+V(\vecr)]\Psi(\vecr)=E\Psi(\vecr)

Since we are only dealing with a $\setminus \vec{r}$ that has a component in $x$, we can use $x$ explicitly:

[(-ħ^2)/(2m)(d^2)/dx^2+V(x)]\Psi(x)=E\Psi(x)

Our question asks us to consider the case where $V \left(x\right) \setminus \equiv V$, independent of any position. Let's also substitute in our wavefunction $\setminus \Psi \left(x\right)$:

[(-ħ^2)/(2m)(d^2)/dx^2+V]Asin(kx)=EAsin(kx)

(-ħ^2)/(2m)(d^2)/dx^2(Asin(kx))+VAsin(kx)=EAsin(kx)

(-ħ^2)/(2m)(d)/dx(kAcos(kx))+VAsin(kx)=EAsin(kx)

(-ħ^2)/(2m)(-k^2Asin(kx))+VAsin(kx)=EAsin(kx)

(ħ^2k^2)/(2m)Asin(kx)+VAsin(kx)=EAsin(kx)

(ħ^2k^2)/(2m)+V=E

Since we have found that the energy associated with this wavefunction is constant and not dependent on position, we have shown the wavefunction $\setminus \Psi \left(x\right)$ satisfies the time-independent Schrödinger equation when $V$ and $E$ are independent of $x$.

It may also be useful to see that ħk is simply momentum $p$. This means we have found:

$E = \frac{{p}^{2}}{2 m} + V$

This makes sense, since energy $E$ is a sum of kinetic energy ${E}_{k}$ and potential energy ${E}_{p}$. In this case, we have the classic form of kinetic energy ${E}_{k} = \frac{{p}^{2}}{2 m}$ and our potential energy is simply a constant ${E}_{p} = V$. This means our solution is valid as it obeys conservation of energy.

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