2
$\begingroup$

I will go straight into an example. Let's take the case of an electron of mass $m$ confined in an infinite 1D box of width $a$. Solving the Schrödinger equation and pay attention to the boundary conditions will give us the set of wave-functions:

$$\psi(x) = \sqrt{\frac{2}{a}}\sin\left( \frac{n\pi}{a} x \right),$$

where $n = 1,2,3...$. The eigenvalues of the Hamiltonian are: $$E = \frac{n^2 \pi^2 \hbar^2}{2ma^2}.$$

Now let's find its de Broglie wavelength. $$ k = \frac{n\pi}{a} \Rightarrow \frac{2\pi}{\lambda} = \frac{n\pi}{a}$$

so, $$\lambda = \frac{2a}{n}.$$ The higher $n$ is, the shorter the wavelength. But what if $n$ is some value like $10000$ or more? The particle has now a very short $\lambda$ compared to the dimension of the system, $a$.

The Question: Does this mean that the electron has now a particle nature more than a wave nature? In other words, is its classical description now "ok"?

Notice that the wave-function above, for $n = 10000$ or more, is still just a sine function and not a localized wave.

I am confused enough. So please, answer only if you are very sure or if you have any references to guide me.

Update:

The states in such a system are discribed by standing waves. The Wikipedia article about "matter waves" says that de Broglie assumed the velocity of the particle to be its group velocity. At least that's what I understand from it. However, a standing wave has $0$ group velocity. That is, the particle is NOT moving in space, and so, it has NO de Broglie wavelength. Now, if $a$ was very big compared to any $\lambda$, then we can assume that the particle can move in "some" space and has some de Broglie wavelength, i. e. it's group velocity is not $0$. The confinement width or system dimensions are the things that play the essential role in this problem.

$\endgroup$

1 Answer 1

1
$\begingroup$

With high energy states, in a small energy band you have enough waves to construct wave packets. Play harmonic composition, here with 40 modes centred around k=50 with a gaussian distribution of amplitudes and the appropriate time exponentials $e^{i k^2 t/2}$ as solutions of the time dependent Schrödinger equation

      Psi[t_, x_] = 
       Evaluate[ Sum[ E^(-1/2 (50. - k)^2) E^(I k^2 /2 t ) Sin[k x],
                 {k, 30, 70, 1}]]

     1.12535*10^-7 E^(1058 I t) Sin[46 x] + 
     0.00012341 E^((2209 I t)/2) Sin[47 x] + 
     0.0183156 E^(1152 I t) Sin[48 x] + 
     0.367879 E^((2401 I t)/2) Sin[49 x] + 
  E^(1250 I t) Sin[50 x] + 
     0.367879 E^((2601 I t)/2) Sin[51 x] + 
     0.0183156 E^(1352 I t) Sin[52 x] + 
     0.00012341 E^((2809 I t)/2) Sin[53 x] + 
     1.12535*10^-7 E^(1458 I t) Sin[54 x]
      

     Manipulate[
      With[{t = Floor[10 ts]/10}, 
        Plot[ Abs[\[Psi][t, x]]^2], {x, -\[Pi], \[Pi]}]], {ts, 0, 2}]

wave packet t=0

wave packet t=0.14

wave packet t=0.35

Conclusion: In a container with reflecting walls, particle like. concentrated wave packets can be constructed at high energies (electrons in the conductain band), but the reflection properties are not amusing. Thats the main reason why solid state physics uses periodic boundary conditions, the flat torus with opposite sides identified and no boundaries.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.