Solution to the 1D Free Schrodinger Equation

[elemis]

So starting from the time dependent Schrödinger equation I perform separation of variables and obtain a time and spatial part. The spatial part is in effect the time independent Schrödinger equation.

Since we are dealing with a free particle I can take the time independent equation, set V = 0 and solve.

I can do this successfully to obtain :

$Ae^{+i\sqrt{{2mE}/{\hbar^{2}}}x}+Be^{-i\sqrt{{2mE}/{\hbar^{2}}}x}$

My lecturer has a small section titled :

>Solving for the Free Schrödinger Equation

$V=0$

$\frac{\hbar^{2}}{2m}\frac{\partial^2\psi}{\partial x^2}+E\psi=0$

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

$\psi=Ce^{-{iEt}/{\hbar}+{ipr}/{\hbar}}$

This is the solution to the free TISE and TDSE.

So when I follow through and solve for the spatial part of his final solution I obtain :

$A'e^{+ip/\hbar}+B'e^{-ip/\hbar}$

He seems to have conveniently ignored/left out one half of the above solution. Why ?

I am also aware that $e^{-{iEt}/{\hbar}}$ is simply the solution the time dependent part of the equation so my only issue is why he has left out a certain bit.

 

[jtbell]

He seems to have conveniently ignored/left out one half of the above solution. Why ?

I don't think anyone here can read his mind. Ask him.

My guess is that he simply did not intend to be completely rigorous.

 

[Chopin]

He's just allowing $p$ to be either positive or negative. Since $E$ is proportional to $p^2$, either positive or negative values will lead to the same value of energy.

 

[jtbell]

Correct. In addition, for the same value of E, both positive and negative values of p can be present, in a superposition, so the two-term solution is a bit more general. We actually use this solution for the "particle in a box" where the boundary conditions lead to coefficients such that the two terms combine to give a (real) sine or cosine.

Of course, that's not a free particle, but it's still V = 0.

 

[Chopin]

Right, but in a free particle with no boundary conditions, you are already allowed to have a superposition of different values of $E$ anyway, so the entire expression is already meant to be taken as only one term in the superposition. If that's the case, it's redundant to also go and write out the positive- and negative- momentum terms separately--you can just say "for any value of $E$, and for any value of $p$ for which $E = \frac{p^2}{2m}$.

 

[AL-Hassan Naser]

please somebody tell us how he canceled out B exp(-ikx) in the spatial part and mention the necessary boundary condition

thx

 

[vanhees71]

He is simply not clear enough about what he wants to do! Obviously he is looking for the energy eigensolutions for a free particle, and then it's clear that you get for each non-zero spectral value of the Hamiltonian two linearly independent generalized eigenfunctions, namely the ones you found. This is called "degeneracy". The reason for a degeneracy is often (if not always) a symmetry of the problem. Here it's invariance under spatial reflections, i.e., the symmetry of the Hamiltonian under $x \rightarrow -x$, $p \rightarrow -p$. Thus, to completely specify the eigenstate you can choose parity as another observable, and this leads to clearly defined eigenfunctions (setting $\hbar=1$):
$$u^{(+)}_E=N \cos(\sqrt{2 m E} x), \quad u^{(-)}_E(x)=N \sin(\sqrt{2m E x}).$$
Of course, also your professor's solution is correct, but it's incomplete, because for any $E>0$ there are two solutions, making up the complete (generalized) basis of the eigenspace $\mathrm{Eig}(\hat{H},E)$.

Another thing is the momentum eigenvalue problem. The momentum spectrum is not degenerate, and then your professor's solution is unique,
$u_p(x)=N \exp(\mathrm{i} p x).$