# Free particle solution to the SE

1. Dec 23, 2015

### DiracPool

The free particle (zero potential) solution of the Schrodinger equation for a plane wave (from my understanding) is ψ= A e^(ikx) + B e^(-ikx). I have 2 questions:

1) This is the so-called free "particle" solution for a plane wave. Does this same solution apply to a light wave/photon also?

2) Are BOTH the A and B expressions involved in describing the wave function? That would describe a wave moving in both directions along the x-axis, wouldn't it? Or would it be a superposition of the two waves, one moving to the left and one moving to the right?

2. Dec 23, 2015

### blue_leaf77

1) The analogy of matter particle like electrons being applied to photons will not work, because the electric field plane wave that people know in classical lightwave, in the view of photon as a quantum object, is expectation value of electric operators in the so-called coherent state of photons. That is, the state of the photon itself is not what we understand as the plane wave in classical optics.

2) The presence of two or more waves in a space must suggest that they are superposing.

3. Dec 23, 2015

### DiracPool

Ok...so are you saying that the free particle solution I listed above applies for an electron? What else? All half-spin particles? I'm not sure what you mean.

Also, if that equation does not apply to photons, what would be the equivalent equation for a propagating photon?

4. Dec 23, 2015

### vanhees71

One should not mix up discussions about photons with non-relativistic quantum mechanics. A photon cannot be appropriately described by a single wave function as a non-relativistic massive particle can.

Now let's consider the time-independent Schrödinger function of a free spinless massive particle. In fact it's nothing else than the eigenvalue problem of the Hamilton operator
$$\hat{H}=\frac{\hat{p}^2}{2m}=-\frac{\hbar^2}{2m} \partial_x^2,$$
assuming that we work about a particle restricted to one dimension.

The eigenvalue equation reads
$$\hat{H} u_E(x)=E u_E(x).$$
You see that for each $E >0$ there are two linearly independent solutions
$$u_E^{(\pm)}(x)=\exp(\pm \mathrm{i} k x), \quad k=\frac{\sqrt{2m E}}{\hbar}.$$
This is understandable: The + solution describes a particle running with a defined momentum $p=\hbar k$ to the right, and the - solution one that moves with the same (magnitude of) momentum to the left. In both cases it has the same energy $E=p^2/(2m)$.

That's called a degeneracy, i.e., for an eigenvalue there are more than one linearly independent eigenvectors. The general eigenstate is given as a superposition of all these eigenvectors.

Last but not least one must emphasize that the plane-wave solutions are not representing true states since they are not square-integrable. They are distributions not true functions in the sense of functional analysis.

5. Dec 23, 2015

### blue_leaf77

It can be applied to all particles having non-relativistic velocity with non-zero rest mass.
I don't think it's a correct view to imagine photon as being a corpuscle-like entity having certain dynamics in space, like it's for electrons for example. From the point of view of quantum optics, the state of photons can take any possible form of excitations of vacuum. You have to start from how lightwave energy is formulated in classical physics, and then use the quantity-to-operator transformation to express observables like electric or magnetic fields as quantum operators. If you further restrict the system to be of empty cavity, you will find that the Hamiltonian of photons takes the form of a quantum harmonics oscillator.
To encompass the case of moving light pulses such as pulsed laser, the closest state of photons which leads to such form of electric field as having sinusoidal oscillation with certain envelope function is the multimode coherent state, this is one of many states photons can have. What we see as moving through the space is the evolution of the expectation value of the electric and magnetic fields, i.e. the expectation values of these operators is time dependent in such a way that it moves through the space, but it's not the photons which are moving in a packet.

6. Dec 26, 2015

### theodoros.mihos

The wave solution $\Psi(x) = A e^{ikx} + B e^{-ikx}$ say nothing about a particle "position". Full solution have the time depend term.
For a classical wave the space is homogen but for a quantum particle the space "formed" by $\Psi(\mathbf{r})$.