# Particle in a Box: Energy Less than the Potential Energy

I am reading quantum mechanics from Shankar's Principles of Quantum Mechanics. On page 157 he defines the box potential $V(x)$ as

$$V(x) = \left\{ \begin{array}{rl} 0 &\mbox{ if |x|< L/2} \\ \infty &\mbox{ otherwise .} \end{array} \right.$$

He starts with region $III$ where $V=\infty$. For this region, he considers $V=V_0>E$ at first and writes the Schrodinger equation as $$\frac{d^2\psi_{III}}{dx^2} + \frac{2m}{\hbar^2} (E-V_0) \psi_{III} = 0.$$

Solving it for $\psi_{III}$, he later shows that $\psi_{III} = 0$ when $V=V_0 \to \infty$.

My confusion is with the assumption: $V=V_0>E$.

Question: How the potential energy $V(x)$ of a particle in region $III$ can be greater than its total energy $E$?

We all know that $E=k_E + V(x)$; $E$ = total energy, $k_E$ = kinetic energy, $V(x)$ = potential energy. If the potential energy is $V_0$, shouldn't the total energy $E$ of the particle be greater than or equal to $V_0$; i.e., $E\geqslant V_0$?

• There is no reason that the potential cannot be larger than energy in quantum physics. It think it would be a good idea to have a textbook on introductory QM which immediately starts with the basic postulates of QM, then some spin-1/2 and then square integrable spaces and only then move on to specific examples like potential wells and the like. – Bubble Oct 6 '14 at 13:17

We all know that $E=k_E + V(x)$; $E$ = total energy, $k_E$ = kinetic energy, $V(x)$ = potential

One reason why you may be confused by this is that the equation $$E = \frac{1}{2}m\dot{x}^2 + V(x)$$ comes from classical mechanics. When we are solving the Schroedinger equation $$H\Phi = E\Phi$$ with Hamiltonian operator $H$, the meaning of the symbols $E$,$H$ is not necessarily the same as in classical mechanics above, even if we call $E$ energy and $H$ Hamiltonian.

Of course, there are similarities in form between the equations of classical mechanics and Schroedinger's equations, but the meaning of the symbols (their use) is different.

We interpret the symbols used in solving the Schroedinger equation with help of the Born interpretation of the function $\Phi$; it gives probability density in configuration space. With this probability density, we can then calculate expected average values of physical quantities, but not their instantaneous values; this is in contrast to classical mechanics, where we deal directly with values, not probabilities.

When we find eigenvalues $E$'s and the corresponding $\Phi$'s, we may use them to make some probabilistic statements about physical quantities of the system considered.

For example, we may calculate average electric moment or average energy:

$$\langle \mu_x \rangle = \int \Phi^* qx \Phi\,dx,$$

$$\langle Energy \rangle = \int \Phi^* H \Phi\,dx,$$

When we put in $\Phi$ corresponding to eigenvalue $E$, we obtain average energy

$$\langle Energy \rangle = E.$$

This shows us possible interpretation of the eigenvalue $E$: the expected average energy of the system appropriate when it is described by $\Phi$. We may say the average energy is $E$, but there is no need to think that instantaneous value of energy is $E$ or that value of energy attached to every point $x$ of the configuration space is $E$. Then, there is no problem with the relation

$$Energy = \frac{1}{2}m\dot{x}^2 + V(x),$$

because the eigenvalue $E$ only gives average energy, not energy corresponding to some point $x$ of configuration space.

Since $k_E \propto \hat{p}^2$ and $\hat{p}$ is Hermitian you may see that this makes $k_E$ positive semidefinite, that is all of its eigenvalues are larger or equal to 0. In other words when you measure this operator you will always get results which are larger or greater than zero. This "contradiction" is resolved by the fact that the potential is a function of the coordinate and the momentum (and hence kinetic energy) does not commute with functions of coordinate. This in turn means that you can't simultaneously measure both $k_E$, $E$, and potential.

• If the momentum (and hence kinetic energy) does not commute with functions of coordinate (the potential energy), then how can we measure the $k_E$ and $V(x)$ both simultaneously? – rainman Oct 6 '14 at 16:13
• Why do you think that we can? – Bubble Oct 6 '14 at 19:31
• Probably I have problem with understanding the following portion of your answer: This "contradiction" is resolved by the fact that the potential is a function of the coordinate and the momentum (and hence kinetic energy) does not commute with functions of coordinate. This in turn means that you can simultaneously measure both $k_E$, $E$, and potential. – rainman Oct 7 '14 at 2:02
• Aha, sorry. Haha. That was a typo. I meant "can't" you can't measure them. That was quite a stupid error on my part. – Bubble Oct 7 '14 at 2:23