# Quantum versus classical mechanics in solids and gases

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.18.

The de Broglie wavelength of a particle, which is the wavelength of an idealized ‘free particle’ which has a precise momentum ${p}$ and thus a completely indeterminate position, is

$\displaystyle \lambda=\frac{h}{p} \ \ \ \ \ (1)$

In general, quantum mechanics is needed to describe systems in which the de Broglie wavelength of the constituent particles is larger than some characteristic size of the system itself. For example, if the wavelength of a free electron (that is, an electron not bound to a particular atom) in a solid is greater than the average spacing between atoms, then quantum mechanics is needed to describe these electrons. If the wavelength is much smaller than the size of the system, the wave nature of a particle isn’t noticeable and we can get away with using classical mechanics.

In statistical mechanics, the average energy of each particle in a system is ${\frac{1}{2}k_{B}T}$ per degree of freedom of the particle, where ${k_{B}}$ is Boltzmann’s constant and ${T}$ is the temperature in kelvins. For a single particle such as an electron, there are three degrees of freedom (one per coordinate direction) so its average energy is

$\displaystyle E=\frac{p^{2}}{2m}=\frac{3}{2}k_{B}T \ \ \ \ \ (2)$

Combining this with the definition of the de Broglie wavelength above, we get

$\displaystyle \lambda=\frac{h}{\sqrt{3mk_{B}T}} \ \ \ \ \ (3)$

so the condition for quantum mechanics to apply is that ${\lambda>d}$ where ${d}$ is the size of the system.

Example 1 Solids. Using the typical lattice spacing of ${d=3\times10^{-10}\mbox{ m}}$ for a solid, what is the maximum temperature at which we need to use quantum mechanics to describe free electrons in such a solid? For quantum mechanics to apply, we need

 $\displaystyle d$ $\displaystyle <$ $\displaystyle \frac{h}{\sqrt{3mk_{B}T}}\ \ \ \ \ (4)$ $\displaystyle T$ $\displaystyle <$ $\displaystyle \frac{h^{2}}{3mk_{B}d^{2}} \ \ \ \ \ (5)$

We have the values (in SI units)

 $\displaystyle h$ $\displaystyle =$ $\displaystyle 6.626\times10^{-34}\ \ \ \ \ (6)$ $\displaystyle m$ $\displaystyle =$ $\displaystyle 9.1\times10^{-31}\ \ \ \ \ (7)$ $\displaystyle k_{B}$ $\displaystyle =$ $\displaystyle 1.38\times10^{-23}\ \ \ \ \ (8)$ $\displaystyle d$ $\displaystyle =$ $\displaystyle 3\times10^{-10} \ \ \ \ \ (9)$

so we get

$\displaystyle T<1.29\times10^{5}\mbox{ K} \ \ \ \ \ (10)$

so electrons most definitely need to be described quantum mechanically.

For the atomic nuclei in a solid, the critical temperature is much lower. For sodium, with an atomic mass of about 23 atomic mass units (where ${1\mbox{ amu}=1.66\times10^{-27}\mbox{ kg}}$), we have

$\displaystyle T<3.1\mbox{ K} \ \ \ \ \ (11)$

Example 2 The ideal gas. The ideal gas law is

$\displaystyle PV=Nk_{B}T \ \ \ \ \ (12)$

where ${P}$ is the pressure, ${V}$ is the volume and ${N}$ is the number of gas molecules. We can get an estimate of ${d}$ by calculating the average volume per molecule ${v}$:

 $\displaystyle v$ $\displaystyle =$ $\displaystyle \frac{V}{N}\ \ \ \ \ (13)$ $\displaystyle d$ $\displaystyle =$ $\displaystyle v^{1/3}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{k_{B}T}{P}\right)^{1/3} \ \ \ \ \ (15)$

Therefore, the condition for quantum mechanics to apply to an ideal gas is

 $\displaystyle \left(\frac{k_{B}T}{P}\right)^{1/3}$ $\displaystyle <$ $\displaystyle \frac{h}{\sqrt{3mk_{B}T}}\ \ \ \ \ (16)$ $\displaystyle T$ $\displaystyle <$ $\displaystyle \frac{h^{6/5}P^{2/5}}{k_{B}\left(3m\right)^{3/5}} \ \ \ \ \ (17)$

For helium at atmospheric pressure we have

 $\displaystyle P$ $\displaystyle =$ $\displaystyle 10^{5}\mbox{N m}^{-2}\ \ \ \ \ (18)$ $\displaystyle m$ $\displaystyle =$ $\displaystyle 4\times\left(1.66\times10^{-27}\right)\mbox{ kg}\ \ \ \ \ (19)$ $\displaystyle T$ $\displaystyle <$ $\displaystyle 2.92\mbox{ K} \ \ \ \ \ (20)$

This is actually below the boiling point of helium (${4.55\mbox{ K}}$) so whenever helium is a gas, we don’t need quantum mechanics to describe it.

For hydrogen atoms (protons) in outer space, ${d=1\mbox{ cm}}$ and ${T=3\mbox{ K}}$. In this case, the critical temperature is given by 5 with ${m=1.66\times10^{-27}\mbox{ kg}}$:

$\displaystyle T<\frac{h^{2}}{3mk_{B}d^{2}}=6.4\times10^{-14}\mbox{ K} \ \ \ \ \ (21)$

Definitely no quantum mechanics needed here.

# Uncertainty principle: an example

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.17.

Here’s another example of calculating the uncertainty principle. We have a wave function defined as

$\displaystyle \Psi\left(x,0\right)=\begin{cases} A\left(a^{2}-x^{2}\right) & -a\le x\le a\\ 0 & \mbox{otherwise} \end{cases} \ \ \ \ \ (1)$

The constant ${A}$ is determined by normalization in the usual way:

 $\displaystyle \int_{-a}^{a}\left|\Psi\right|^{2}dx$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A^{2}\left.\left(\frac{x^{5}}{5}-\frac{2}{3}a^{2}x^{3}+a^{4}x\right)\right|_{-a}^{a}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A^{2}\frac{16a^{5}}{15}\ \ \ \ \ (4)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{\sqrt{15}}{4a^{5/2}} \ \ \ \ \ (5)$

The expectation value of ${x}$ is ${\left\langle x\right\rangle =0}$ from the symmetry of the wave function. The expectation value of ${p}$ is

 $\displaystyle \left\langle p\right\rangle$ $\displaystyle =$ $\displaystyle -i\hbar\int_{-a}^{a}\Psi^*\frac{\partial}{\partial x}\Psi dx\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\int_{-a}^{a}\left(-2Ax\right)A\left(a^{2}-x^{2}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (8)$

[We can’t calculate ${\left\langle p\right\rangle =\frac{d}{dt}\left(m\left\langle x\right\rangle \right)}$ in this case, because we know the value of ${\left\langle x\right\rangle }$ only at one specific time (${t=0}$), so we don’t have enough information to calculate its derivative.]

The remaining statistics are (the integrals are all just integrals of polynomials, so nothing complicated):

 $\displaystyle \left\langle x^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-a}^{a}x^{2}\left|\Psi\right|^{2}dx\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{15}{16a^{5}}\left.\left(\frac{x^{7}}{7}-\frac{2}{5}a^{2}x^{5}+\frac{1}{3}a^{4}x^{3}\right)\right|_{-a}^{a}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{a^{2}}{7}\ \ \ \ \ (11)$ $\displaystyle \left\langle p^{2}\right\rangle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\int_{-a}^{a}\Psi^*\frac{\partial^{2}}{\partial x^{2}}\Psi dx\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{15\hbar^{2}}{8a^{5}}\left.\left(a^{2}-x^{2}\right)\right|_{-a}^{a}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{5\hbar^{2}}{2a^{2}}\ \ \ \ \ (14)$ $\displaystyle \sigma_{x}$ $\displaystyle =$ $\displaystyle \sqrt{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{a}{\sqrt{7}}\ \ \ \ \ (16)$ $\displaystyle \sigma_{p}$ $\displaystyle =$ $\displaystyle \sqrt{\left\langle p^{2}\right\rangle -\left\langle p\right\rangle ^{2}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{5}{2}}\frac{\hbar}{a}\ \ \ \ \ (18)$ $\displaystyle \sigma_{x}\sigma_{p}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{5}{14}}\hbar\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle \cong$ $\displaystyle 0.598\hbar>\frac{\hbar}{2} \ \ \ \ \ (20)$

Thus the uncertainty principle is satisfied in this case.

# Inner product of two wave functions is constant in time

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.16.

The fact that the normalization of the wave function is constant over time is actually a special case of a more general theorem, which is

$\displaystyle \frac{d}{dt}\int_{-\infty}^{\infty}\Psi_{1}^*\Psi_{2}dx=0 \ \ \ \ \ (1)$

for any two normalizable solutions to the Schrödinger equation (with the same potential). The proof of this follows a similar derivation to that in section 1.4 of Griffiths’s book.

The derivative in the integrand is (where we’re using a subscript ${t}$ or ${x}$ to denote a derivative with respect that variable):

 $\displaystyle \frac{\partial}{\partial t}\left(\Psi_{1}^*\Psi_{2}\right)$ $\displaystyle =$ $\displaystyle \Psi_{1t}^*\Psi_{2}+\Psi_{1}^*\Psi_{2t} \ \ \ \ \ (2)$

From the Schrödinger equation

 $\displaystyle \Psi_{2t}$ $\displaystyle =$ $\displaystyle i\frac{\hbar}{2m}\Psi_{2xx}-\frac{i}{\hbar}V\Psi_{2}\ \ \ \ \ (3)$ $\displaystyle \Psi_{1t}^*$ $\displaystyle =$ $\displaystyle -i\frac{\hbar}{2m}\Psi_{1xx}^*+\frac{i}{\hbar}V\Psi_{1}^*\ \ \ \ \ (4)$ $\displaystyle \Psi_{1t}^*\Psi_{2}+\Psi_{1}^*\Psi_{2t}$ $\displaystyle =$ $\displaystyle i\frac{\hbar}{2m}\left(-\Psi_{1xx}^*\Psi_{2}+\Psi_{2xx}\Psi_{1}^*\right)+\frac{i}{\hbar}V\left(\Psi_{1}^*\Psi_{2}-\Psi_{1}^*\Psi_{2}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\frac{\hbar}{2m}\frac{\partial}{\partial x}\left(\Psi_{2x}\Psi_{1}^*-\Psi_{1x}^*\Psi_{2}\right) \ \ \ \ \ (6)$

Inserting this into 1 and integrating gives zero because all wave functions go to zero at infinity. [Of course, the theorem doesn’t hold if ${\Psi_{1}}$ and ${\Psi_{2}}$ are solutions for different potentials, because in that case the potential term wouldn’t cancel out in 5.]

# Unstable particles: a crude model

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.15.

A rather unrealistic way of modelling an unstable particle is to introduce an imaginary component to the potential. We can see this by modifying the proof given in Griffiths’s section 1.4 that, for a real potential, the normalization of the wave function is constant in time. We propose that

$\displaystyle V\left(x\right)=V_{0}\left(x\right)-i\Gamma \ \ \ \ \ (1)$

where ${V_{0}}$ is the ‘true’ potential and ${\Gamma}$ is a positive real constant.

The Schrödinger equation then says (where we’re using a subscript ${t}$ or ${x}$ to denote a derivative with respect that variable):

 $\displaystyle i\hbar\Psi_{t}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\Psi_{xx}+V_{0}\Psi-i\Gamma\Psi\ \ \ \ \ (2)$ $\displaystyle -i\hbar\Psi_{t}^*$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\Psi_{xx}^*+V_{0}\Psi^*+i\Gamma\Psi^* \ \ \ \ \ (3)$

where the second line is the complex conjugate of the first.

Retaining the interpretation of the wave function as a probability of finding the particle at a given place and time, we can calculate the time derivative of the total probability of finding the particle anywhere:

 $\displaystyle \frac{dP}{dt}\equiv\frac{d}{dt}\int_{-\infty}^{\infty}\left|\Psi\right|^{2}dx$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\frac{\partial}{\partial t}\left|\Psi\right|^{2}dx \ \ \ \ \ (4)$

The derivative in the integrand is

 $\displaystyle \frac{\partial}{\partial t}\left|\Psi\right|^{2}$ $\displaystyle =$ $\displaystyle \frac{\partial}{\partial t}\left(\Psi^*\Psi\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Psi_{t}^*\Psi+\Psi^*\Psi_{t} \ \ \ \ \ (6)$

From 2 and 3 we have

 $\displaystyle \Psi_{t}$ $\displaystyle =$ $\displaystyle i\frac{\hbar}{2m}\Psi_{xx}-\frac{i}{\hbar}V_{0}\Psi-\frac{\Gamma}{\hbar}\Psi\ \ \ \ \ (7)$ $\displaystyle \Psi_{t}^*$ $\displaystyle =$ $\displaystyle -i\frac{\hbar}{2m}\Psi_{xx}^*+\frac{i}{\hbar}V_{0}\Psi^*-\frac{\Gamma}{\hbar}\Psi^*\ \ \ \ \ (8)$ $\displaystyle \Psi_{t}^*\Psi+\Psi^*\Psi_{t}$ $\displaystyle =$ $\displaystyle i\frac{\hbar}{2m}\left(\Psi_{xx}\Psi^*-\Psi_{xx}^*\Psi\right)-2\frac{\Gamma}{\hbar}\Psi^*\Psi\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\frac{\hbar}{2m}\frac{\partial}{\partial x}\left(\Psi_{x}\Psi^*-\Psi_{x}^*\Psi\right)-2\frac{\Gamma}{\hbar}\Psi^*\Psi \ \ \ \ \ (10)$

Putting this into 4 we can integrate the first term and get zero because ${\Psi\rightarrow0}$ as ${x\rightarrow\pm\infty}$ so we’re left with

$\displaystyle \frac{dP}{dt}=-2\frac{\Gamma}{\hbar}\int_{-\infty}^{\infty}\left|\Psi\right|^{2}dx=-2\frac{\Gamma}{\hbar}P \ \ \ \ \ (11)$

We can solve this differential equation to get

 $\displaystyle \frac{dP}{P}$ $\displaystyle =$ $\displaystyle -2\frac{\Gamma}{\hbar}dt\ \ \ \ \ (12)$ $\displaystyle P$ $\displaystyle =$ $\displaystyle P_{0}e^{-2\Gamma t/\hbar} \ \ \ \ \ (13)$

where ${P_{0}}$ is the probability of finding the particle at ${t=0}$. If we know the particle hasn’t decayed at ${t=0}$ then ${P_{0}=1}$.

The half-life of the particle is the time it takes for ${P}$ to be reduced to ${P_{0}/2}$, so

 $\displaystyle \ln0.5$ $\displaystyle =$ $\displaystyle -2\frac{\Gamma}{\hbar}t_{1/2}\ \ \ \ \ (14)$ $\displaystyle t_{1/2}$ $\displaystyle =$ $\displaystyle 0.347\frac{\hbar}{\Gamma} \ \ \ \ \ (15)$