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Suppose there is a rectangular solid, with dimensions $l_x$, $l_y$ and $l_z$, and imagine that an electron inside experiences no force at all, except at the impenetrable walls:

$$V(x,y,z)= \begin{array}{cc} \{ & \begin{array}{cc} 0, & 0<x<l_x, 0<y<l_y, 0<z<l_z \\ \infty, & otherwise \\ \end{array} \end{array} $$

The (normalized) wave functions are

$$\psi_{n_xn_yn_z}=\sqrt{\frac{8}{l_xl_yl_z}}sin(\frac{n_x\pi}{l_x}x)sin(\frac{n_y\pi}{l_y}y)sin(\frac{n_z\pi}{l_z}z),$$

and the allowed energies are

$$E_{n_xn_yn_z}=\frac{\hbar^2\pi^2}{2m}(\frac{n_x^2}{l_x^2}+\frac{n_y^2}{l_y^2}+\frac{n_z^2}{l_z^2})=\frac{\hbar^2k^2}{2m},$$

where $k$ is the magnitude of the $\textbf{wave vector, k}\equiv(k_x,k_y,k_z)$.

My questions are:

  1. The wave function(considering the part of time also) is not a travelling wave in case of infinite potential well so what's the significance of wave vector in this case? Its real and imaginary parts are standing waves but as a whole can I say that it's a standing wave?

  2. If I solve $E=\frac{\hbar^2k^2}{2m}$ using $m=\frac{E}{c^2}$, $E=h\nu$ and $\frac{c}{\nu}=\lambda$, I get $k=\frac{2\sqrt{2}\pi}{\lambda}$ while in fact k should be equal to $\frac{2\pi}{\lambda}$. Where am I mistaken?

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  • $\begingroup$ Equation really is $E^2=m^2c^4+p^2c^2$. But I don't understand what do you want to achieve. $\endgroup$
    – Jitendra
    Aug 26, 2018 at 7:38
  • $\begingroup$ Considering ques 1. As far as I know, wave vector is in the direction of wave propagation in a lossless isotropic medium and is defined for only travelling waves. That's why I want to ask whether the given wave function is a travelling wave because k is called the wave vector. $\endgroup$
    – Asit Srivastava
    Aug 26, 2018 at 7:57
  • $\begingroup$ It's a standing wave, which is formed by two exactly same travelling waves but propagating in opposite directions. $\endgroup$
    – Jitendra
    Aug 26, 2018 at 9:17
  • $\begingroup$ How can it be proved from $E=\frac{\hbar^2k^2}{2m}$ that $k=\frac{2\pi}{\lambda}$? $\endgroup$
    – Asit Srivastava
    Aug 26, 2018 at 10:11
  • $\begingroup$ You do not need $E=\frac{\hbar^2k^2}{2m}$ to prove $k=\frac{2\pi}{\lambda}$. Time period of an oscillation is defined as $T=\frac{\lambda}{c}$ iff velocity is $c$, otherwise following on @okruz's answer as velocity here really is $\frac{2\pi\nu}{k}$, $T=\frac{\lambda k}{2\pi\nu}$, which gives $k=\frac{2\pi}{\lambda}$ on using $T=\frac{1}{\nu}$. $\endgroup$
    – Jitendra
    Aug 26, 2018 at 11:02

1 Answer 1

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The formula $E=mc^2$ is a relativistic one and does not really make sense in the context of your solution to the "particle in a box" problem which assumes a constant mass.

Secondly (given that $c$ refers to the velocity of light in vacuum), the formula $\frac{c}{\nu}=\lambda$ assumes that the phase velocity of the wave is $c$, which is not the case. The phase velocity is $c_{p}=\frac{\omega}{k} = \frac{E}{\hbar k} = \frac{\hbar k}{2m}$ (with $\omega = 2\pi\nu$).

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