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May 11, 2024

Rayleigh-Jeans law

The Rayleigh-Jeans law is a flawed attempt in physics to describe the spectral radiance of electromagnetic radiation as a function of wavelength from a blackbody at a given temperature.

Consider a cube of side $L$ , filled with electromagnetic radiation in thermal equilibrium at temperature $T$ . If a tiny aperture exists in one of the walls, the radiation that it emits would exhibit the properties of an ideal blackbody. The walls, regarded as perfect conductors, provide boundary conditions for the electromagnetic waves inside the cube.

In a perfect conductor, an idealised model for real metals, electric charges can move without any resistance. An electromagnetic wave propagating in the $x$ -direction, with its electric field vector parallel to the wall, causes electrons in the conductor to move in response to the field. The movement of these electrons creates an opposing electric field that cancels out the external electric field. Given that the electric field is zero at the walls, the boundary conditions can only be satisfied by standing waves. These are waves that oscillate in place and have nodes at the cube’s walls. For example, standing waves oscillating in the $x$ -direction have wavelengths of ${\lambda_x=2L/n_x}$ , where ${n_x}$ is an integer (see below diagram).

The amplitude $A$ of each of the above standing waves can be expressed as ${A=A_0sin\frac{2\pi x}{\lambda_x}}$ , or in terms of the wave vector ${\boldsymbol{\mathit{k}}_x}$ pointing in the $x$ -direction:

$A=A_0sin$\boldsymbol{\mathit{k}}_x\cdot\boldsymbol{\mathit{x}}$=A_0sink_xx$

where ${\vert\boldsymbol{\mathit{k}}_x\vert=k_x=\frac{2\pi}{\lambda_x}=\frac{\pi n_x}{L}}$ .

Similarly, the wavelengths of the oscillation modes of the electromagnetic field in the $y$ -direction and $z$ -direction are ${\lambda_y=2L/n_y}$ and ${\lambda_z=2L/n_z}$ respectively. Correspondingly, the magnitudes of the wave vectors are given by ${k_y=\pi n_y/L}$ and ${k_z=\pi n_z/L}$ respectively. It follows that the wave vector $\boldsymbol{\mathit{k}}$ of the oscillation modes of the electromagnetic field in an arbitrary direction in the cube lies in a $k$ -space, which contain a cubical array of points that are $\pi/L$ apart (see Figure I below).

Since the magnitude of the wave vector $\boldsymbol{\mathit{k}}$ is ${\vert\boldsymbol{\mathit{k}}\vert=k=\sqrt{k_x^2+k_y^2+k_z^2}=\frac{\pi}{L}\sqrt{n_x^2+n_y^2+n_z^2}}$ , boundary conditions are satisfied by each ordered triple ${(k_x,k_y,k_z)}$ . In other words, every mode of oscillation of an arbitrary wave oscillating in the cube is defined by a point in the $k$ -space. We can also plot the array of modes in an $n$ -space, where the points are now unit length apart from one another (see Figure II above). Substituting ${k=2\pi/\lambda=2\pi\nu/c}$ in ${k=\frac{\pi}{L}\sqrt{n_x^2+n_y^2+n_z^2}}$ , we have

$\frac{2L}{c}\nu=\sqrt{n_x^2+n_y^2+n_z^2}=R\;\;\;\;\;\;\;\;1$

where $R$ is the radius of an octant (one-eighth of the volume of a sphere) in the $n$ -space.

For a certain value of $\nu$ , ordered triples of ${(n_x,n_y,n_z)}$ that satisfy eq1 lie on the surface of the octant. Consider the volume of the shell of the octant ${dV=\frac{1}{8}4\pi R^2dR}$ enclosed by the surfaces at $R$ and $R+dR$ . Substituting eq1 and ${\frac{2L}{c}d\nu=dR}$ in $d\nu$ , we have

$dV=4\pi\biggr$\frac{L}{c}\biggr$^3\nu^2d\nu$

Since the density $D$ of the number of modes in the $n$ -space is one per unit volume, the number of modes $Nd\nu$ within the shell is the product of $D$ and $dV$ :

$Nd\nu=4\pi\biggr$\frac{L}{c}\biggr$^3\nu^2d\nu$

As there are two independent polarisations for each mode of an electromagnetic wave (see above diagram),

$Nd\nu=8\pi\biggr$\frac{L}{c}\biggr$^3\nu^2d\nu$

Lord Rayleigh used the equipartition theorem to assume that each mode has an average energy of $\overline{E}=kT$ . Multiplying the above equation by the average energy $\overline{E}$ per mode, we have

$u(\nu)d\nu=8\pi\frac{\nu^2}{c^3}\overline{E}d\nu=\frac{8\pi\nu^2kT}{c^3}d\nu\;\;\;\;\;\;\;\;2$

where ${u(\nu)d\nu=\frac{N\overline{E}}{L^3}d\nu}$ is the energy density (average energy per mode between $\nu$ and $\nu+d\nu$ per unit volume of the cube).

Eq2 is known as the Rayleigh-Jeans law, which fails to describe the spectrum of blackbody radiation at high frequencies because it implies that radiated energy would increase without limit as $\nu$ increases, leading to what is called an ultraviolet catastrophe (see diagram below).

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May 11, 2024June 16, 2024

Planck radiation law

The Planck radiation law explains how a blackbody emits electromagnetic radiation at a specific temperature, based on the assumption that the energy of each oscillator in the body can only have discrete values.

In June 1900, Lord Rayleigh published the Rayleigh-Jeans law, which is now known as a flawed attempt in physics to describe the spectral radiance of electromagnetic radiation as a function of wavelength from a blackbody at a given temperature. The mistake that he made was to use the equipartition theorem to assume that each oscillation mode within a blackbody has an average energy of $\overline{E}=kT$ . In December of the same year, the German physicist Max Planck presented the Planck radiation law, which assumed that the energy of an oscillator of frequency $\nu$ came in discrete bundles:

$E_{n,\nu}=nh\nu$

where $n=0,1,2,\cdots$ and $h$ is a proportionality constant called the Planck constant.

According to the Boltzmann distribution, the probability of a mode $p_{n,\nu}$ with frequency $\nu$ associated with the state $E_{n,\nu}$ is

$p_{n,\nu}=\frac{e^{-\frac{E_{n,\nu}}{kT}}}{\sum_{n=0}^{\infty}e^{-\frac{E_{n,\nu}}{kT}}}$

The average energy $\overline{E}$ of the mode of frequency $\nu$ is

$\overline{E}=\sum_{n=0}^{\infty}E_{n,\nu}p_{n,\nu}=\frac{\sum_{n=0}^{\infty}E_{n,\nu}e^{-\frac{E_{n,\nu}}{kT}}}{\sum_{n=0}^{\infty}e^{-\frac{E_{n,\nu}}{kT}}}=h\nu\frac{\sum_{n=0}^{\infty}ne^{-\frac{nh\nu}{kT}}}{\sum_{n=0}^{\infty}e^{-\frac{n\nu}{kT}}}$

Let ${x=e^{-\frac{h\nu}{kT}}}$ .

$\overline{E}=h\nu\frac{\sum_{n=0}^{\infty}nx^n}{\sum_{n=0}^{\infty}x^n}=h\nu\biggr$\frac{x+2x^2+3x^3+\cdots}{1+x+x^2+\cdots}\biggr$=h\nu x\biggr$\frac{1+2x+3x^2+\cdots}{1+x+x^2+\cdots}\biggr$$

Substituting the Taylor series of ${\frac{1}{1-x}=1+x+x^2+\cdots}$ and ${\frac{1}{(1-x)^2}=1+2x+3x^2+\cdots}$ in the above equation,

$\overline{E}=\frac{h\nu x}{1-x}=\frac{h\nu}{x^{-1}-1}=\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}\;\;\;\;\;\;\;\;4$

Substituting eq4 in eq2,

$u(\nu)d\nu=\frac{8\pi h\nu^3}{c^3}\frac{1}{e^{\frac{h\nu}{kT}}-1}d\nu\;\;\;\;\;\;\;\;5$

which is the mathematical expression of the Planck distribution law.

Question

Show that in the classical limit, the average energy of a mode in eq4 is consistent with the equipartition theorem.

Answer

In the classical limit, ${h\nu\rightarrow 0}$ and we can expand ${e^\frac{h\nu}{kT}}$ as the Taylor series ${e^\frac{h\nu}{kT}=1+\frac{h\nu}{kT}+\frac{1}{2!}\biggr$\frac{h\nu}{kT}\biggr$^2+\cdots}$ . Substituting the series in eq4 and ignoring the higher powers of the series because ${\frac{h\nu}{kT}\rightarrow 0}$ , we have $\overline{E}=kT$ .

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May 9, 2024May 10, 2024

Maxwell-Boltzmann distribution

The Maxwell-Boltzmann distribution $f(v)$ describes the range of speeds $v$ among gaseous molecules of an ideal gas at a given temperature $T$ .

The energy of an ideal gas molecule of mass $m$ is characterised solely by its kinetic energy $E$ . For a molecule with velocity $\hat{v}$ , its energy is ${E=\frac{1}{2}mv_x^2+\frac{1}{2}mv_y^2+\frac{1}{2}mv_z^2}$ , where ${v_x}$ , ${v_y}$ and ${v_z}$ are the velocity vector components of $\hat{v}$ in Cartesian coordinates. Given that a molecule with the velocity magnitude ${\vert\hat{v}_i\vert=v_i}$ has the same energy regardless of direction, any molecule with the ordered triple ${(v_x,v_y,v_z)}$ that lies on the surface of a sphere of radius ${v_i}$ in velocity space belongs to the energy state ${\varepsilon_i=\frac{1}{2}mv_{x,i}^2+\frac{1}{2}mv_{y,i}^2+\frac{1}{2}mv_{,i}z^2}$ (see diagram below).

The fraction of molecules in the energy state ${\varepsilon_i}$ is given by the Boltzmann distribution:

$\frac{n_i}{N}=\frac{e^{-\frac{\varepsilon_i}{kT}}}{\sum_ie^{-\frac{\varepsilon_i}{kT}}}\;\;\;\;\;\;\;\;1$

where

$k$ is the Boltzmann constant.
$T$ is the equilibrium temperature of the system.
${n_i}$ is the number of molecules in the energy state.
$N$ is the total number of molecules in the system.

In probability theory, ${\frac{n_i}{N}}$ is the probability ${p_i}$ of molecules in state $i$ , with ${\sum_ie^{-\frac{\varepsilon_i}{kT}}}$ being the normalisation constant. In other words, ${p_i\propto e^{-\frac{\varepsilon_i}{kT}}}$ , which we can rewrite as:

$f(v_x,v_y,v_z)=Ke^{-\frac{m(v_x^2+v_y^2+v_z^2)}{2kT}}\;\;\;\;\;\;\;\;2$

where ${f(v_x,v_y,v_z)}$ is the probability density function (probability per unit interval of $v$ ) of molecules in a particular energy state ${\frac{1}{2}mv_x^2+\frac{1}{2}mv_y^2+\frac{1}{2}mv_z^2}$ (we have dropped the index $i$ for convenience) and $K$ is the proportionality constant.

Since ${f(v_x,v_y,v_z)=\biggr$K'e^{-\frac{mv_x^2}{2kT}}\biggr$\biggr$K'e^{-\frac{mv_y^2}{2kT}}\biggr$\biggr$K'e^{-\frac{mv_z^2}{2kT}}\biggr$=f(v_x)f(v_y)f(v_z)}$ , where ${K'=K^{1/3}}$ ,

$f(v_x,v_y,v_z)=Ke^{-\frac{mv_x^2}{2kT}}e^{-\frac{mv_y^2}{2kT}}e^{-\frac{mv_z^2}{2kT}}\;\;\;\;\;\;\;\;3$

As the probability density function ${f(v_x,v_y,v_z)}$ is separable, its normalisation is equivalent to the separate normalisation of each of the individual functions:

$\int_{-\infty}^{\infty}f(v_x)dv_x=1\;\;\;\;\int_{-\infty}^{\infty}f(v_y)dv_y=1\;\;\;\;\int_{-\infty}^{\infty}f(v_z)dv_z=1\;\;\;\;\;\;\;\;4$

Let’s evaluate just one of the integrals in eq4 by substituting ${f(v_x)=K^{1/3}e^{-\frac{mv_x^2}{2kT}}}$ in ${\int_{-\infty}^{\infty}f(v_x)dv_x=1}$ to give

$K^{\frac{1}{3}}\int_{-\infty}^{\infty}e^{-\frac{mv_x^2}{2kT}}dv_x=1\;\;\;\;\;\;\;\;5$

Question

Show that ${\int_{-\infty}^{\infty}e^{-aX^2}dX=\sqrt{\frac{\pi}{a}}}$ .

Answer

Let ${I=\int_{-\infty}^{\infty}e^{-aX^2}dX}$ and at the same time let ${I=\int_{-\infty}^{\infty}e^{-aY^2}dY}$ since $X$ and $Y$ are dummy variables.

$I^2=\int\int_{-\infty}^{\infty}e^{-a(X^2+Y^2)}dXdY$

In polar coordinates, $r^2=X^2+Y^2,X=rcos\theta,Y=rsin\theta$ . For infinitesimal changes, the change in area $dA$ can be approximated as a rectangle with sides $r$ and $rd\theta$ (see diagram below), with $dXdY=dA=rd\theta dr$ .

So,

$I^2=\int_{0}^{2\pi}d\theta\int_{0}^{\infty}re^{-r^2}dr=2\pi\int_{0}^{\infty}re^{-r^2}dr$

Let $u=r^2\;\Rightarrow\;du=2rdr$

$I^2=2\pi\int_{0}^{\infty}re^{-au}\frac{du}{2r}=\pi\int_{0}^{\infty}e^{-au}du=\frac{\pi}{a}\;\;\Rightarrow\;\;I=\sqrt{\frac{\pi}{a}}$

Hence, eq5 becomes,

$K=\biggr$\frac{m}{2\pi kT}\biggr$^{\frac{3}{2}}\;\;\;\;\;\;\;\;6$

Substituting eq6 in eq3,

$f(v_x,v_y,v_z)=\biggr$\frac{m}{2\pi kT}\biggr$^{\frac{3}{2}}e^{-\frac{mv_x^2}{2kT}}e^{-\frac{mv_y^2}{2kT}}e^{-\frac{mv_z^2}{2kT}}\;\;\;\;\;\;\;\;7$

Eq7 is known as the Maxwell-Boltzmann distribution for the velocity vector of a molecule. The fraction of molecules ${dn_{v_xv_yv_z}/N}$ with velocity vectors in the infinitesimal range of ${v_x}$ to ${v_x+dv_x}$ , ${v_y}$ to ${v_y+dv_y}$ and ${v_z}$ to ${v_z+dv_z}$ is

$\frac{dn_{v_xv_yv_z}}{N}=f(v_x,v_y,v_z)dv_xdv_ydv_z=\biggr$\frac{m}{2\pi kT}\biggr$^{\frac{3}{2}}e^{-\frac{m(v_x^2+v_y^2+v_z^2)}{2kT}}dv_xdv_ydv_z\;\;\;\;\;\;\;\;8$

The fraction of molecules ${dn_v/N}$ with speed in the infinitesimal range of $v$ to $v+dv$ is then equivalent to the fraction of molecules with velocity vectors that lies within a spherical shell of infinitesimal thickness in velocity space (see diagram below).

In other words,

$\frac{dn_v}{N}=\biggr$\frac{m}{2\pi kT}\biggr$^{\frac{3}{2}}e^{-\frac{mv^2}{2kT}}\sum_{shell}dv_xdv_ydv_z$

where ${v^2=v_x^2+v_y^2+v_z^2}$ , and we have assumed that the volume of the shell of infinitesimal thickness is approximately equal to the sum of infinitesimal cubes, each with an infinitesimal volume of ${dv_xdv_ydv_z}$ .

Therefore, ${\sum_{shell}dv_xdv_ydv_z=4\pi v^2dv}$ and ${\frac{dn_v}{N}=f(v)dv=4\pi\biggr$\frac{m}{2\pi kT}\biggr$^{\frac{3}{2}}v^2e^{-\frac{mv^2}{2kT}}dv}$ ., which implies that the probability density function for the speed of a molecule $f(v)$ is

$f(v)=4\pi\biggr$\frac{m}{2\pi kT}\biggr$^{\frac{3}{2}}v^2e^{-\frac{mv^2}{2kT}}\;\;\;\;\;\;\;\;9$

Eq9 is known as the Maxwell-Boltzmann distribution for the speed of a molecule. It also expresses the probability of a molecule being associated with a particular energy state. For a mole of an ideal gas, eq9 becomes

$f(v)=4\pi\biggr$\frac{M}{2\pi RT}\biggr$^{\frac{3}{2}}v^2e^{-\frac{Mv^2}{2RT}}\;\;\;\;\;\;\;\;10$

where $M$ is the mass of one mole of the gas and $R$ is the gas constant.

Question

Show that the most probable speed of one mole of an ideal gas is ${v=\sqrt{\frac{2RT}{M}}}$ .

Answer

The most probable speed corresponds to the peak of the distribution (see above diagram). Carrying out the derivative ${\frac{df(v)}{dv}}$ , equating it to zero, and noting that the most probable speed $v$ is neither $0$ nor $\infty$ , we have ${v=\sqrt{\frac{2RT}{M}}}$ .

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May 8, 2024

Statistical entropy

Statistical entropy is a concept in thermodynamics that quantifies the number of different ways a system can be arranged while still maintaining its macroscopic properties.

The total energy of a system, $E$ , is given by eq5. Its total differential is:

$dE=\sum_i\varepsilon_idn_i+\sum_in_id\varepsilon_i$

Consider the transfer of heat to a closed system of particles. If the volume of the system is constant, then no work is done on the system and the energy level of each state is unchanged $d\varepsilon_i=0$ . However, the population of particles in each state $dn_i$ varies according to temperature. Hence,

$dE=\sum_i\varepsilon_idn_i\;\;\;\;\;\;\;\;26$

If we interpret $E$ as the internal energy of the system, $U$ , then according to the first law of thermodynamics of a constant volume system,

$dE=dU=dq\;\;\;\;\;\;\;\;27$

For a reversible transfer of heat, the second law of thermodynamics states that ${dS=\frac{dq}{T}}$ . Combining $dS$ , eq26 and eq27,

$dS=k\beta\sum_i\varepsilon_idn_i\;\;\;\;\;\;\;\;28$

Substituting eq16 in eq28,

$dS=k\sum_i\biggr$\frac{\partial lnW}{\partial n_i}\biggr$dn_i+k\alpha\sum_idn_i$

Since the total number of particles for a closed system is constant, ${\sum_idn_i=0}$ and

$dS=k\sum_i\biggr$\frac{\partial lnW}{\partial n_i}\biggr$dn_i\;\;\;\;\;\;\;\;29$

Substituting eq4 in eq29, we have $dS=kdlnW$ or the integrated form:

$S=klnW\;\;\;\;\;\;\;\;30$

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May 8, 2024July 3, 2024

Boltzmann distribution

The Boltzmann distribution, formulated in 1868 by Ludwig Boltzmann, describes the probability distribution $p_i$ of objects (particles or oscillation modes) in a system over various energy states, $\varepsilon_i$ .

It is mathematically expressed as:

$p_i=\frac{n_i}{N}=\frac{e^{-\frac{\varepsilon_i}{kT}}}{\sum_ie^{-\frac{\varepsilon_i}{kT}}$

where

$k$ is the Boltzmann constant.
$n_i$ is the number of objects in the energy state $\varepsilon_i$ .
$N$ is the total number of objects in the system.

The derivation of the Boltzmann distribution equation involves the following steps:

1. Derivation of the total differential of $lnW$ .
2. Application of the Lagrange method of undetermined multipliers on the total differential of $lnW$ .
3. Simplification of solution using Stirling’s approximation.
4. Evaluation of $\beta$ .

Step 1

Consider a system with molecules randomly occupying different energy states, $\varepsilon_i$ . At any time, the configuration of the system can be represented by $\{n_0,n_1,\cdots\}$ with $n_0$ molecules in energy state $\varepsilon_0$ , $n_1$ molecules in energy state $\varepsilon_1$ , and so on. The total number of molecules is therefore:

$\sum_in_i=N\;\;\;\;\;\;\;\;i=0,1,\cdots\;\;\;\;\;\;\;\;1$

where $n_i$ is the number of molecules in the energy state $\varepsilon_i$ .

The number of ways, $W$ , to achieve an instantaneous configuration of $\{n_0,n_1,\cdots\}$ is given by the combinatorial mathematics of

$W=\frac{N!}{n_0!n_1!\cdots}\;\;\;\;\;\;\;\;2$

or in the natural logarithmic form:

$lnW=ln\frac{N!}{n_0!n_1!\cdots}\;\;\;\;\;\;\;\;3$

Question

Eq2 implies that the molecules are distinguishable. Why?

Answer

Boltzmann statistics is rooted in classical physics, where particles of the same type are considered distinguishable because they can be differentiated by their physical states, such as position and velocity. In contrast, particles of the same type in quantum mechanics are considered indistinguishable, which leads to different statistical distributions like Fermi-Dirac statistics for fermions and Bose-Einstein statistics for bosons.

As the number of molecules in each energy state $\{n_0,n_1,\cdots\}$ varies with time, the configuration of the system changes and so does the number of ways of achieving the new configurations. We can therefore express the LHS of eq3 in its total differential form of
${dlnW=\sum_i\biggr$\frac{\partial lnW}{\partial n_i}\biggr$_{all\;other\;n\neq n_i}dn_i}$ , or for simplicity:

$dlnW=\sum_i\biggr$\frac{\partial lnW}{\partial n_i}\biggr$dn_i\;\;\;\;\;\;\;\;4$

Let’s further define our system as a closed system with total energy, $E$ , given by:

$\sum_in_i\varepsilon_i=E\;\;\;\;\;\;\;\;i=0,1,\cdots\;\;\;\;\;\;\;\;5$

Eq5 restricts the number of configurations of the system. For example, the configurations of $\{N,0,0,\cdots\}$ and $\{N-1,1,0,\cdots\}$ cannot coexist as they have different total energies. If we assume, under the conditions imposed by eq1 and eq5, that all possible configurations of the system have the same probability of occurring, the configuration with the maximum number of ways of achieving will most likely be the one the system adopts.

Step 2

The most probable configuration is found by evaluating the maximum point for the function in eq4, i.e.

$dlnW=\sum_i\biggr$\frac{\partial lnW}{\partial n_i}\biggr$dn_i=0\;\;\;\;\;\;\;\;6$

To solve eq 6, we employ the Lagrange method of undetermined multipliers. We begin by rearranging eq1 to $n_j=N-n_0-n_1-\cdots-n_{j-1}-n_{j+1}-\cdots$ , where $n_j$ is dependent on the rest of the variables, which are all independent. Eq1 can also be written in the form of a new function, :

$g=n_0+n_1+\cdots-N=0\;\;\;\;\;\;\;\;7$

Likewise, by rearranging eq5 to ${n_k=\frac{1}{\varepsilon_k}(E-n_0\varepsilon_0-n_1\varepsilon_1-\cdots-n_{k-1}\varepsilon_{k-1}-\cdots)}$
we have another dependent variable, $n_k$ and another function:

$h=n_0\varepsilon_0+n_1\varepsilon_1+\cdots-E=0\;\;\;\;\;\;\;\;8$

This results in eq6 having two dependent variables. The total differential of $g$ and $h$ are

$dg=\sum_i\biggr$\frac{\partial g}{\partial n_i}\biggr$dn_i=0\;\;\;\;\;\;\;\;9$

and

$dh=\sum_i\varepsilon_i\biggr$\frac{\partial h}{\partial n_i}\biggr$dn_i=0\;\;\;\;\;\;\;\;10$

respectively.

Since $dg=dh=0$ , we can multiply eq9 and eq10 by the factors $\alpha$ and $-\beta$ respectively and add them to eq6 to give:

$\sum_i\biggr\[\biggr$\frac{\partial lnW}{\partial n_i}\biggr$+\alpha\biggr$\frac{\partial g}{\partial n_i}\biggr$-\beta\varepsilon_i\biggr$\frac{\partial h}{\partial n_i}\biggr$\biggr\]dn_i=0\;\;\;\;\;\;\;\;11$

The factors, $\alpha$ and $\beta$ , are called Lagrange multipliers. Two of the variables, e.g. $n_j$ and $n_k$ , are dependent variables, while the rest are independent variables. If there is some value of $\alpha$ and some value of $\beta$ that render the $j$ -th and $k$ -th terms of eq11 zero, we have

$\biggr$\frac{\partial lnW}{\partial n_j}\biggr$+\alpha\biggr$\frac{\partial g}{\partial n_j}\biggr$-\beta\varepsilon_j\biggr$\frac{\partial h}{\partial n_j}\biggr$=0\;\;\;\;\;\;\;\;12$

$\biggr$\frac{\partial lnW}{\partial n_k}\biggr$+\alpha\biggr$\frac{\partial g}{\partial n_k}\biggr$-\beta\varepsilon_k\biggr$\frac{\partial h}{\partial n_k}\biggr$=0\;\;\;\;\;\;\;\;13$

Consequently, we are left with all independent variables terms. $dn_i$ in eq11 can now vary arbitrarily, which implies that all the remaining coefficients equal to zero. Substituting eq9 and eq10 in eq11,

$\sum_i\biggr$\frac{\partial lnW}{\partial n_i}\biggr$dn_i+\alpha dg-\beta dh=0\;\;\;\;\;\;\;\;14$

Noting that $N$ and $E$ are constants, eq7 and eq8 become ${dg=dn_0+dn_1+\cdots=\sum_idn_i}$ and ${dh=dn_0\varepsilon_0+dn_1\varepsilon_1+\cdots=\sum_i\varepsilon_idn_i}$ respectively. Substituting $dg$ and $dh$ in eq14,

$\sum_i\biggr\[\biggr$\frac{\partial lnW}{\partial n_i}\biggr$+\alpha-\beta\varepsilon_i\biggr\] dn_i=0\;\;\;\;\;\;\;\;15$

Since all coefficients are now equal to zero,

$\biggr$\frac{\partial lnW}{\partial n_i}\biggr$+\alpha-\beta\varepsilon_i=0\;\;\;\;\;\;\;\;16$

Step 3

To simplify eq16, we take the natural logarithm on both sides of eq3 to give ${lnW=lnN!-\sum_iln(n_i!)}$ . Since ${lnN!=ln[N(N-1)\cdots]=lnN+ln(N-1)+\cdots=\sum_{K=1}^NlnK}$ ,

$lnW=\sum_{K=1}^NlnK-\sum_iln(n!)\;\;\;\;\;\;\;\;17$

For large $N$ , we have ${\sum_{K=1}^NlnK\approx\int_1^NlnKdK}$ . Integrating by parts, ${\int_1^NlnKdK=NlnN-N}$ . Hence, ${lnN!=\sum_{K=1}^NlnK\approx NlnN-N}$ , which is known as Stirling’s approximation. Eq17 becomes ${lnW=NlnN-N-\biggr\[\sum_in_ilnn_i-\sum_in_i\biggr\]}$ . Since, ${\sum_in_i=N}$ ,

$lnW=NlnN-\sum_in_ilnn_i\;\;\;\;\;\;\;\;18$

Substituting eq18 in eq16,

$lnN\frac{\partial N}{\partial n_i}+N\frac{\partial lnN}{\partial n_i}-\sum_j\frac{\partial n_jlnn_j}{\partial n_i}+\alpha-\beta\varepsilon_i=0\;\;\;\;\;\;\;\;19$

where we have changed the summation index from $i$ to $j$ in eq19 to discriminate the summation variable from the differentiation variable.

Since $N=n_0+n_1+\cdots$ , we have ${\frac{\partial N}{\partial n_i}=1}$ . By implicit differentiation, ${\frac{\partial lnN}{\partial n_i}=\frac{1}{N}\frac{\partial N}{\partial n_i}=\frac{1}{N}}$ and ${\frac{\partial lnn_j}{\partial n_i}=\frac{1}{n_j}\frac{\partial n_j}{\partial n_i}}$ . Furthermore, ${\frac{\partial n_j}{\partial n_i}=\delta_{ij}$ . Therefore, eq19 becomes,

$n_i=Ne^{\alpha}e^{-\beta\varepsilon_i}\;\;\;\;\;\;\;\;20$

Substituting eq20 in eq1, we have ${e^{\alpha}=1/\sum_ie^{-\beta\varepsilon_i}}$ , which when substituted in eq20 gives

$\frac{n_i}{N}=\frac{e^{-\beta\varepsilon_i}}{\sum_ie^{-\beta\varepsilon_i}}\;\;\;\;\;\;\;\;21$

Step 4

An easy way to determine the value of $\beta$ is to use the equation for the distribution of molecules of an ideal gas in a cylinder:

$\frac{n}{n_0}=e^{-\frac{mg\Delta h}{kT}}\;\;\;\;\;\;\;\;22$

where

$n$ is number of molecules at a height $l$ , which implies that $n$ is number of molecules in energy state $\varepsilon_l$ .
$n_0$ is the number of molecules at a height $j$ , where $j < l$ . It follows that $n_0$ is number of molecules in energy state $\varepsilon_j$ .
$m$ is the mass of a molecule.
$g$ is the acceleration due to gravity.
$\Delta h$ is the difference in height between $l$ and $j$ .

Since $mg\Delta h$ represents the difference in energy between states $l$ and $j$ , we can rewrite eq22 as:

$\frac{n_l}{n_j}=e^{-\frac{\varepsilon_l-\varepsilon_j}{kT}}\;\;\;\;\;\;\;\;23$

From eq21, the fractions of molecules in energy states $l$ and $j$ are ${\frac{n_l}{N}=\frac{e^{-\beta\varepsilon_l}}{\sum_ie^{-\beta\varepsilon_i}}}$ and ${\frac{n_j}{N}=\frac{e^{-\beta\varepsilon_j}}{\sum_ie^{-\beta\varepsilon_i}}}$ respectively. Dividing ${\frac{n_l}{N}}$ by ${\frac{n_j}{N}}$ ,

$\frac{n_l}{n_j}=e^{-\beta(\varepsilon_l-\varepsilon_j)}\;\;\;\;\;\;\;\;24$

Comparing eq23 and 24, $\beta=1/kT$ . Therefore, eq21 becomes

$\frac{n_i}{N}=\frac{e^{-\frac{\varepsilon_i}{kT}}}{\sum_ie^{-\frac{\varepsilon_i}{kT}}}\;\;\;\;\;\;\;\;25$

which is the Boltzmann distribution.

The Boltzmann distribution is used to derive mathematical expressions of many scientific concepts, including the statistical entropy, the Maxwell-Boltzmann distribution, and the Planck radiation law.

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April 30, 2024

The harmonic oscillator

A harmonic oscillator comprises a particle that is subject to a restoring force proportional to the particle’s displacment $x$ from its equilibrium position.

Consider a mass $m$ connected to a rigid support by a massless and frictionless spring (see above diagram). At equilibrium, the length of the spring is $x_0$ . Let’s assume that the only force $F$ acting on the mass is a restoring force that is directly proportional to the displacement of the mass (Hooke’s law):

$F=-kx\;\;\;\;\;\;\;\;1$

where $x=x_1-x_0$ is the displacement of the mass from its equilibrium position and $k$ is constant of proportionality called the force constant.

Question

Explain why $k$ is a measure of the stiffness of the spring.

Answer

For a particular value of $F$ , the stiffer the spring, the less the mass displaces. Since $k$ is inversely proportional to $x$ , it is a measure of the stiffness of the spring.

Substituting the equation for Newton’s 2^nd law of motion $F=ma$ in eq1, we have

$m\frac{d^2x}{dt^2}+kx=0\;\;\;\;\;\;\;\;2$

The solution to the above differential equation is $x=Asin\omega t$ , where $\omega=\sqrt{\frac{k}{m}}$ . As the displacement of the mass is defined by a wave equation, it is called a harmonic oscillator (‘harmonic’ originates from sound waves). The mass oscillates with amplitude $A$ and frequency $\frac{\omega}{2\pi}$ (see Q&A below), and the kinetic energy $E_k$ and potential energy $U$ of the system are

$E_k=\frac{1}{2}mv^2=\frac{1}{2}m\biggr$\frac{dx}{dt}\biggr$^2=\frac{1}{2}m$A\omega cos\omega t$^2$

$U=\frac{1}{2}kx^2=\frac{1}{2}k$Asin\omega t$^2\;\;\;\;\;\;\;\;3$

Question

Explain why the mass oscillates with a frequency of $\frac{\omega}{2\pi}$ and why the potential energy of the system is equal to $\frac{1}{2}kx^2$ ?

Answer

Since $x=Asin\omega t=Asin\biggr\[\omega\biggr$t+\frac{2\pi}{\omega}\biggr$\biggr\]$ , the function repeats itself after a time $\frac{2\pi}{\omega}$ . This implies that the period $T$ of the motion of the mass is $T=\frac{2\pi}{\omega}$ and that the oscillation frequency $\nu$ is $\nu=\frac{1}{T}=\frac{\omega}{2\pi}$ .

The force exerted by a person $F_p$ in lifting an object over a distance $x$ against gravity is $F_p=\frac{W}{\Delta x}$ , where $W$ is the work done by the person. $W$ must also be the amount of potential energy the object gains, i.e. $F_p=\frac{\Delta U}{\Delta x}$ . Furthermore, the force exerted by gravity $F$ is opposite to the force exerted by the person. So, $F=-\frac{\Delta U}{\Delta x}$ and for small changes, $F=-\frac{dU}{dx}$ . Since gravitational force and elastic spring force are both conservative forces (work done is path-independent, we substitute eq1 in $F=-\frac{dU}{dx}$ to give $dU=kxdx$ . Therefore, the potential energy of a harmonic oscillator is $U=k\int_0^xxdx=\frac{1}{2}kx^2$ .

Therefore, the total energy $E$ of the harmonic oscillator is

$E=\frac{1}{2}m$A\omega cos\omega t$^2+\frac{1}{2}k$Asin\omega t$^2=\frac{1}{2}kA^2$

and the Hamiltonian $H$ , which is the sum of the kinetic and potential energies of the oscillator can be expressed as

$H=\frac{p^2}{2m}+\frac{1}{2}kx^2$

Before we show how the harmonic oscillator can be used to model a vibrating diatomic molecule, we shall look at the quantum-mechanical treatment of a harmonic oscillator.

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April 30, 2024

Generating function and Rodrigues’ formula for the Hermite polynomials

The generating function $g(y,t)$ for Hermite polynomials is a mathematical tool that, when expanded as a power series, yields the ratio of the Hermite polynomials $H_n(y)$ and the factorial of the summation index as its coefficients.

It can also be used to derive the normalisation constant for the wavefunctions of the quantum harmonic oscillator, and is defined as

$g(y,t)=e^{-t^2+2ty}=\sum_{n=0}^{\infty}\frac{H_n(y)}{n!}t^n=H_0(y)+H_1(y)t+\frac{1}{2!}H_2(y)t^2+\cdots\;\;\;\;\;\;\;\;33$

To prove the 2^nd equality of eq33, we take the partial derivative of $g$ with respect to $y$ to give ${\frac{\partial g(y,t)}{\partial y}=\sum_{n=0}^{\infty}\frac{dH_n(y)}{dy}\frac{t^n}{n!}}$ . Substituting eq29 in this equation, we have

${\frac{\partial g(y,t)}{\partial y}=\sum_{n=0}^{\infty}2nH_{n-1}(y)\frac{t^n}{n!}=2H_0(y)t+\frac{2(2)}{2!}H_1(y)t^2+\frac{2(3)}{3!}H_2(y)t^3+\cdots=2tg(y,t)}$

The solution to the differential equation ${\frac{\partial g(y,t)}{\partial y}-2tg(y,t)=0$ is

$g(y,t)=e^{2ty}f(t)\;\;\;\;\;\;\;\;34$

When $y=0$ , eq34 becomes

$g(0,t)=f(t)=\sum_{n=0}^{\infty}\frac{H_n(0)}{n!}t^n=H_0(0)+H_1(0)t+\frac{H_2(0)}{2!}t^2+\cdots$

Using eq19,

$f(t)=H_0(0)+\frac{H_2(0)}{2!}t^2+\frac{H_4(0)}{4!}t^4+\cdots=\sum_{n=0}^{\infty}\frac{H_{2n}(0)}{(2n)!}t^{2n}$

Using eq20,

$f(t)=\sum_{n=0}^{\infty}\frac{(-1)^n(2n)!}{(2n)!n!}t^{2n}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}t^{2n}$

Since the Maclaurin series expansion of $e^x$ is ${\sum_{n=0}^{\infty}\frac{x^n}{n!}}$ , we have ${f(t)=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}t^{2n}=e^{-t^2}}$ , which completes the proof.

The generating function can be used to derive the Rodrigues formula for Hermite polynomials. From 33,

$g(y,t)=e^{-t^2+2ty}=e^{y^2-(t-y)^2}=e^{y^2}e^{-(t-y)^2}$

$\frac{\partial g(y,t)}{\partial t}=e^{y^2}\frac{\partial}{\partial t}e^{-(t-y)^2}\;\;\;\;\;\;\;\;35$

Let $f=e^{-(t-y)^2$ . We have ${\frac{\partial f}{\partial t}=-(2t-2y)e^{-(t-y)^2}}$ and ${\frac{\partial f}{\partial y}=(2t-2y)e^{-(t-y)^2}}$ . Therefore,

$\frac{\partial f}{\partial t}=-\frac{\partial f}{\partial y}\;\;\;\;\;\;\;\;36$

Substitute eq36 in eq35

$\frac{\partial g(y,t)}{\partial t}=e^{y^2}\frac{\partial}{\partial t}e^{-(t-y)^2}=(-1)e^{y^2}\frac{\partial}{\partial y}e^{-(t-y)^2}$

$\frac{\partial^n g(y,t)}{\partial t^n}=e^{y^2}\frac{\partial^n}{\partial t^n}e^{-(t-y)^2}=(-1)e^{y^2}\frac{\partial}{\partial y}\frac{\partial^{n-1}}{\partial t^{n-1}}e^{-(t-y)^2}=\\(-1)^ne^{y^2}\frac{\partial^n}{\partial y^n}e^{-(t-y)^2}\;\;\;\;\;\;\;n=1,2,\cdots\;\;\;\;\;\;\;\;37$

When $t=0$

$\frac{\partial^n g(y,t)}{\partial t^n}_{t=0}=(-1)^ne^{y^2}\frac{\partial^n}{\partial y^n}e^{-y^2}\;\;\;\;\;\;\;\;38$

From eq33

$\frac{\partial g(y,t)}{\partial t}=H_1(y)+H_2(y)t+\frac{1}{2}H_3(y)t^2+\frac{1}{6}H_4(y)t^3\cdots$

$\frac{\partial^2 g(y,t)}{\partial t^2}=H_2(y)+H_3(y)t+\frac{1}{2}H_4(y)t^2+\cdots$

$\frac{\partial^n g(y,t)}{\partial t^n}=H_n(y)+H_{n+1}(y)t+\frac{1}{2}H_{n+2}(y)t^2+\cdots$

$\frac{\partial^n g(y,t)}{\partial t^n}_{t=0}=H_n(y)\;\;\;\;\;\;\;\;39$

Substitute eq39 into eq38

$H_n(y)=(-1)^ne^{y^2}\frac{d^n}{dy^n}e^{-y^2}\;\;\;\;\;\;\;\;40$

Eq40 is the Rodrigues formula for Hermite polynomials. In other words, the Rodrigues formula for Hermite polynomials is a mathematical expression that provides a method to calculate any Hermite polynomial using differentiation.

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Taylor series of single-variable and multi-variable functions

The Taylor series is a way to express a function as an infinite sum of terms, each of which is derived from the function’s derivative value at a specific point $a$ .

Consider the power series $f(x)=c_0+c_1(x-a)+c_2(x-a)^2+\cdots+c_n(x-a)^n$ . We have $f(a)=c_0$ , $f'(a)=c_1$ , $f''(a)=2c_2$ , $f'''(a)=6c_3$ and so on. Therefore,

$f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots+\frac{f^n(a)}{n!}(x-a)^n\;\;\;\;\;\;\;\;32b$

Eq32b is called a Taylor series, which approximates a function in the neighbourhood of the point $a$ . When $a=0$ , the Taylor series is also known as a Maclaurin series:

$f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\cdots+\frac{f^n(0)}{n!}x^n$

The input value $x$ of $f$ in eq32b represents points in the domain of $f$ that are near $a$ . In other words, we can express $f(x)$ as $f(a+t(x-a))$ , where $a$ is a constant, $t$ is a variable, and $x-a$ represents a small change in $a$ that is scaled by $t$ . This alternate expression of $f$ is useful when dealing with multiple-variable functions.

For a multivariable function $f(x_1,x_2,\cdots,x_n)$ , the Taylor series about the point $(a_1,a_2,\cdots,a_n)$ is

$f(x_1,x_2,\cdots,x_n)=f(a_1,a_2,\cdots,a_n)+\sum_{i=1}^n\frac{\partial f(a_1,a_2,\cdots,a_n)}{\partial x_i}(x_i-a_i)+\\\frac{1}{2!}\frac{\partial^2 f(a_1,a_2,\cdots,a_n)}{\partial x_i\partial x_j}(x_i-a_i)(x_j-a_j)+\cdots\;\;\;\;\;\;\;\;32c$

Question

How do we use vector notation to express the function $f(x_1,x_2,\cdots,x_n)=c_1x_1+c_2x_2+\cdots+c_nx_n$ ?

Answer

If we let $\hat{x}=(x_1,x_2,\cdots,x_n)$ , then we write $f(\hat{x})$ in place of $f(x_1,x_2,\cdots,x_n)$ . In other words, we have $f(\hat{x})=\hat{c}\cdot\hat{x}$ , where $\hat{c}=(c_1,c_2,\cdots,c_n)$ . Consequently, $f$ can be regarded as

1. a function of $n$ variables $x_1,x_2,\cdots,x_n$ ,
2. a function of a single point variable $(x_1,x_2,\cdots,x_n)$ , or
3. a function of a single vector variable $\hat{x}$ .

To derive eq32c, let $\hat{x}=(x_1,x_2,\cdots,x_n)$ and $\hat{a}=(a_1,a_2,\cdots,a_n)$ . Consider the function

$g(t)=f(\hat{b}=\hat{a}+t(\hat{x}-\hat{a}))\;\;\;\;\;\;\;\;32d$

which implies that $\hat{b}=(x_1,x_2,\cdots,x_n)$ .

$f(\hat{a}+t(\hat{x}-\hat{a}))$ is a multivariable function, whose input is the vector $\hat{a}+t(\hat{x}-\hat{a})$ , which varies with $t$ . In the domain of $f(x_1,x_2,\cdots,x_n)$ , $\hat{a}$ is the vector representing a point where $f$ is expanded, and $(\hat{x}-\hat{a})$ is a fixed vector that determines the direction of the displacement from the point $\hat{a}$ . In other words, $(\hat{x}-\hat{a})$ is not a variable of $g(t)$ , but a parameter that we choose before plotting $g(t)$ . The function $g(t)$ is therefore a single variable function of $t$ , and its points are of the form $(t,g(t))$ .

The Maclaurin series expansion of $g(t)$ when $t=1$ is

$g(1)=g(0)+\frac{g'(0)}{1!}+\frac{g''(0)}{2!}+\cdots\;\;\;\;\;\;\;\;32e$

According to the multivariable chain rule, ${\frac{df(\hat{b})}{dt}=\frac{\partial f(\hat{b})}{\partial x_1}\frac{dx_1}{dt}+\frac{\partial f(\hat{b})}{\partial x_2}\frac{dx_2}{dt}+\cdots+\frac{\partial f(\hat{b})}{\partial x_n}\frac{dx_n}{dt}=\nabla f(\hat{b})\cdot\frac{d\hat{x}}{dt}}$ , where ${\nabla=\boldsymbol{\mathit{i}}\frac{\partial}{\partial x_1}+\boldsymbol{\mathit{j}}\frac{\partial}{\partial x_2}+\cdots+\boldsymbol{\mathit{k}}\frac{\partial}{\partial x_n}}$ and $\hat{x}=\boldsymbol{\mathit{i}}x_1+\boldsymbol{\mathit{j}}x_2+\cdots+\boldsymbol{\mathit{k}}x_n$ . So,

$g'(t)=\nabla f(\hat{b})\cdot\frac{d\hat{x}}{dt}=\cdots=\sum_{i=1}^n\frac{\partial f(\hat{b})}{\partial x_i}(x_i-a_i)\;\;\;\;\;\;\;\;32f$

The second derivative $g''(t)=f''(\hat{b})$ is

$g''(t)=\frac{d^2f(\hat{b})}{dt^2}=\frac{d}{dt}\biggr\[\frac{\partial f(\hat{b})}{\partial x_1}\frac{dx_1}{dt}+\frac{\partial f(\hat{b})}{\partial x_2}\frac{dx_2}{dt}+\cdots+\frac{\partial f(\hat{b})}{\partial x_n}\frac{dx_n}{dt}\biggr\]$

Using the multivariable chain rule again, we get

$g''(t)=\frac{d^2f(\hat{b})}{dt^2}=\sum_{i,j=1}^n\frac{\partial^2 f(\hat{b})}{\partial x_i\partial x_j}\frac{dx_i}{dt}\frac{dx_j}{dt}+\sum_{k=1}^n\frac{\partial f(\hat{b})}{\partial x_k}\frac{d^2x_k}{dt^2}\;\;\;\;\;\;\;\;32g$

With reference to eq32d, $\hat{b}=\hat{a}+t(\hat{x}-\hat{a})$ with components $x_i=a_i+t(x_i-a_i)$ . It follows that $\frac{dx_i}{dt}=x_i-a_i$ , $\frac{dx_j}{dt}=x_j-a_j$ and $\frac{d^2x_k}{dt^2}=0$ . Eq32g becomes

$g''(t)=\frac{d^2f(\hat{b})}{dt^2}=\sum_{i,j=1}^n\frac{\partial^2 f(\hat{b})}{\partial x_i\partial x_j}(x_i-a_i)(x_j-a_j)\;\;\;\;\;\;\;\;32h$

Furthermore, when $t=0$ , we have $\hat{b}=\hat{a}$ . So, eq32f and eq32h become

$g'(0)=\sum_{i=1}^n\frac{\partial f(\hat{a})}{\partial x_i}(x_i-a_i)\;\;\;\;\;\;\;\;32i$

and

$g''(0)=\sum_{i,j=1}^n\frac{\partial^2 f(\hat{a})}{\partial x_i\partial x_j}(x_i-a_i)(x_j-a_j)\;\;\;\;\;\;\;\;32j$

respectively.

Substituting eq32i and eq32j in eq32e and noting that $g(0)=f(\hat{a})$ and $g(1)=f(\hat{x})$ , we have eq32c. Finally, the Maclaurin series of a multivariable function is

$\begin{matrix}f(x_1,x_2,\cdots,x_n)=\\\\\end{matrix}\begin{matrix}f(a_1=0,a_2=0,\cdots,a_n=0)+\\\\\sum_{i=1}^n\frac{\partial f(a_1=0,a_2=0,\cdots,a_n=0)}{\partial x_i}x_i+\\\\\frac{1}{2!}\frac{\partial^2 f(a_1=0,a_2=0,\cdots,a_n=0)}{\partial x_i\partial x_j}x_ix_j+\cdots\end{matrix}\;\;\;\;\;\;\;\;32k$

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The quantum harmonic oscillator

The quantum harmonic oscillator is a model for studying an atom that moves back and forth about an equilibrium point. The model’s central premise involves the use of the potential energy term of the classical harmonic oscillator to construct the Schrodinger equation for the harmonic oscillator:

$\biggr$\frac{p^2}{2m}+\frac{1}{2}kx^2\biggr$\psi(x)=E\psi(x)\;\;\;\;\;\;\;\;4$
which is equivalent to

$\biggr$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}kx^2\biggr$\psi(x)=E\psi(x)$

where $x$ is the displacement of the mass from its equilibrium position.

Question

Show that $\frac{p^2}{2m}$ is equivalent to $-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$ .

Answer

The de Broglie relation is $p=\hbar k$ , where $k=\frac{2\pi}{\lambda}$ . The derivative of the wavefunction $\psi=Ae^{ikx}$ with respect to $x$ is

$\frac{d\psi}{dx}=ikAe^{ikx}=ik\psi$

$\frac{\hbar}{\sqrt{2m}}\frac{d}{dx}\psi=i\frac{\hbar}{\sqrt{2m}}k\psi=i\frac{p}{\sqrt{2m}}\psi$

This suggests that the operator $\frac{\hbar}{\sqrt{2m}}\frac{d}{dx}$ is equivalent to $i\frac{p}{\sqrt{2m}}$ , and hence, $\hat{p}=\frac{\hbar}{i}\frac{d}{dx}$ . Squaring the momentum operator $\hat{p}$ and dividing by $2m$ gives $-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$ .

Substituting $k=m\omega^2$ in the above equation and rearranging, we have

$\frac{d^2}{dx^2}\psi(x)+\biggr$\frac{2m}{\hbar^2}E-\frac{m^2\omega^2}{\hbar^2}\biggr$\psi(x)=0\;\;\;\;\;\;\;\;5$

The probability of locating an electron in the Hilbert space must be finite, i.e. $\int_{-\infty}^{\infty}\vert\psi(x)\vert^2dx<\infty$ . This implies that $\psi(x)$ vanishes as $x\rightarrow\pm\infty$ . We call such a well-behaved wavefunction square-integrable. Hence, we need to study the asymptotic characteristics of eq5 to find possible solutions.

Let’s begin by simplifying the differential equation using a change of variable. Substituting $x=y\biggr$\frac{\hbar}{m\omega}\biggr$^{1/2}$ and $\epsilon=\frac{E}{\hbar\omega}$ in eq5, we have

$\frac{d^2\psi(y)}{dy^2}+\(2\epsilon-y^2)\psi(y)=0\;\;\;\;\;\;\;\;6$

Since $2\epsilon$ is a constant, $y^2\gg 2\epsilon$ as $y\rightarrow\pm\infty$ , and eq6 approximates to

$\frac{d^2\psi(y)}{dy^2}-y^2\psi(y)=0\;\;\;\;\;\;\;\;7$

This suggests that the solution to eq6 could be Gaussian function because $\psi(y)=e^{-\frac{y^2}{2}}$ is a solution to eq7. Let’s try $\psi(y)=u(y)e^{-\frac{y^2}{2}}$ , where $u(y)$ is a function of $y$ . Substituting $\psi(y)=u(y)e^{-\frac{y^2}{2}}$ in eq6 and computing the derivatives, we have

$\frac{d^2u(y)}{dy^2}-2y\frac{du(y)}{dy}+(2\epsilon-1)u(y)=0\;\;\;\;\;\;\;\;8$

Eq8 is known as the Hermite differential equation. Since any well-behaved function can be expressed as a power series, let $u(y)=\sum_{n=0}^{\infty}C_ny^n$ and substitute $\frac{du(y)}{dy}=\sum_{n=0}^{\infty}nC_ny^{n-1}$ and $\frac{d^2u(y)}{dy^2}=\sum_{n=0}^{\infty}n(n-1)C_ny^{n-2}$ in eq8 to give

$\sum_{n=0}^{\infty}n(n-1)C_ny^{n-2}-2y\sum_{n=0}^{\infty}nC_ny^{n-1}+(2\epsilon-1)\sum_{n=0}^{\infty}C_ny^{n}=0$

Substituting $2y\sum_{n=0}^{\infty}nC_ny^{n-1}=2\sum_{n=0}^{\infty}nC_ny^n$ and $\sum_{n=0}^{\infty}n(n-1)C_ny^{n-2}=\sum_{n=0}^{\infty}(n+2)(n+1)C_{n+2}y^n$ in the above equation, we have

$\sum_{n=0}^{\infty}\[(n+2)(n+1)C_{n+2}-2nC_n+(2\epsilon-1)C_n\]y^n=0$

The above equation is only true for all values of $y$ if every coefficient for each power of $y$ is zero. So, $(n+2)(n+1)C_{n+2}-2nC_n+(2\epsilon-1)C_n=0$ , which rearranges to

$C_{n+2}=\frac{2n+1-2\epsilon}{(n+2)(n+1)}C_n\;\;\;\;\;\;\;\;9$

Eq9 is a recurrence relation. If we know the value of $C_0$ , we can use the relation to find $C_2,C_4,C_6,\cdots$ . Similarly, if we know $C_1$ , we can find $C_3,C_5,C_7,\cdots$ .

Comparing the recurrence relations for even-labelled coefficients,

$C_{2n}=-\[2\epsilon-1-2(2n-2)\]\frac{(2n-2)!}{(2n)!}C_{2n-2}\;\;\;where\;n=1,2,3\cdots\;\;\;\;10$

Similarly, the recurrence relations for odd-labelled coefficients can be expressed as

$C_{2n+1}=-\[2\epsilon-3-2(2n-2)\]\frac{(2n-1)!}{(2n+1)!}C_{2n-1}\;\;\;where\;n=1,2,3\cdots\;\;\;\;11$

Rewriting $\psi(y)$ as a sum of two power series, one containing terms with even-labelled coefficients and the other with odd-labelled coefficients,

$\psi(y)=\biggr\{C_0+\sum_{l=1}^{\infty}\biggr\[-\[2\epsilon-1-2(2l-2)\]\frac{C_{2l-2}}{\frac{(2l)!}{(2l-2)!}}\biggr\]y^{2l}\biggr\}e^{-\frac{y^2}{2}}+\\\biggr\{C_1y+\sum_{l=1}^{\infty}\biggr\[-\[2\epsilon-3-2(2l-2)\]\frac{C_{2l-1}}{\frac{(2l+1)!}{(2l-1)!}}\biggr\]y^{2l+1}\biggr\}e^{-\frac{y^2}{2}}\;\;\;\;\;\;\;\;12$

According to L’Hopital’s rule, $\psi(y)$ appears to be a possible solution to eq7 because each of the terms in eq12 (simplified to $Cy^n/e^{\frac{y^n}{2}}$ ) approaches zero for large $y$ :

$\lim\limits_{y\rightarrow\pm\infty}\frac{Cy^n}{e^{\frac{y^2}{2}}}=\lim\limits_{y\rightarrow\pm\infty}\frac{nCy^{n-1}}{e^{\frac{y^2}{2}}}=\cdots=\lim\limits_{y\rightarrow\pm\infty}\frac{k}{e^{\frac{y^2}{2}}}=0$

where $k$ is a constant.

However, this does not guarantee that the series as a whole also converges because the sum of an infinite number of infinitesimal terms can still be infinite, as is the case for a harmonic series. To see how the two infinite series behave for large $y$ , we carry out the ratio test as follows:

where we have set $n=2l$ and $n=2l+1$ in eq9 for the even series and odd series respectively.

By comparison, the ratio test for the power series expansion of $e^{y^2}=\sum_{l=0}^{\infty}\frac{y^{2l}}{l!}$ is also $\frac{1}{l}$ . Therefore, both infinite series behave the same as $e^{y^2}$ and diverge for large $y$ . To ensure that $\psi(y)$ is square-integrable, we need to truncate either one of the series after some finite terms $l$ and let all the coefficients of the other series be zero.

Question

Can we instead truncate both series of eq12 after some $l$ ?

Answer

Since the value of $l$ is arbitrary, the sum of two truncated series, each with finite terms, may still have an infinite number of terms. The only way to guarantee that $u(y)$ has finite terms is to truncate one of the series and let $C_0$ , if the odd series is truncated, or $C_1$ , if the even series is truncated, be zero.

To truncate either series, we let the numerator of eq9 be zero so that every successive term in the selected series is zero as well. This implies that

$\epsilon=n+\frac{1}{2}\;\;\;\;\;\;\;\;13$

Let’s rewrite eq12 as

$\psi(y)=\biggr\{\begin{matrix}\biggr\{C_0+\sum_{l=1}^{m}\biggr\[-\[2\epsilon-1-2(2l-2)\]\frac{C_{2l-2}}{\frac{(2l)!}{(2l-2)!}}\biggr\]y^{2l}\biggr\}e^{-\frac{y^2}{2}}&even\;series\;l=1,2,\cdots\\\biggr\{C_1y+\sum_{l=1}^{m}\biggr\[-\[2\epsilon-3-2(2l-2)\]\frac{C_{2l-1}}{\frac{(2l+1)!}{(2l-1)!}}\biggr\]y^{2l+1}\biggr\}e^{-\frac{y^2}{2}}&odd\;series\;l=1,2,\cdots\end{matrix}\;\;\;\;14$

Since eq14 is the result of substituting $\psi(y)=u(y)e^{-\frac{y^2}{2}}$ in eq6, it is a solution to eq6.

Question

Show that $\psi_4(y)=C_0\biggr\[1-\frac{2n}{2!}y^2+\frac{2^2n(n-2)}{4!}y^4\biggr\]e^{-\frac{y^2}{2}}$ .

Answer

We expand the even series in eq14 as follows:

$\psi_4(y)=\biggr\[C_0-(2\epsilon-1)\frac{C_0}{2!}y^2-2(2\epsilon-1-4)\frac{C_2}{4!}y^4\biggr\]e^{-\frac{y^2}{2}}$

Substituting eq9, where $C_2=\frac{1-2\epsilon}{2}C_0$ , in the above equation, we have

$\psi_4(y)=\biggr\[C_0-(2\epsilon-1)\frac{C_0}{2!}y^2-\frac{(2\epsilon-1-4)(1-2\epsilon)}{4!}C_0y^4\biggr\]e^{-\frac{y^2}{2}}$

Substituting eq13 in the above equation gives the required expression.

Re-expanding the even series of eq14 using eq9 and eq13, we have

$\boldsymbol{\mathit{n}}$	$\boldsymbol{\mathit{\psi_n(y)}}$
$0$	$C_0e^{-\frac{y^2}{2}}$
$2$	$C_0\biggr\[1-\frac{2n}{2!}y^2\biggr\]e^{-\frac{y^2}{2}}$
$4$	$C_0\biggr\[1-\frac{2n}{2!}y^2+\frac{2^2n(n-2)}{4!}y^4\biggr\]e^{-\frac{y^2}{2}}$
$6$	$C_0\biggr\[1-\frac{2n}{2!}y^2+\frac{2^2n(n-2)}{4!}y^4-\frac{2^3n(n-2)(n-4)}{6!}y^6\biggr\]e^{-\frac{y^2}{2}}$

Comparing the coefficients of each even series in the above table, we have

$\psi_n(y)=C_0\biggr\[1-\frac{2n}{2!}y^2+\frac{2^2n(n-2)}{4!}y^4+\cdots+\frac{2^n$\frac{n}{2}$!}{(-1)^{\frac{n}{2}}n!}y^n\biggr\]e^{-\frac{y^2}{2}}\;\;\;\;\;\;\;\;15a$

where $n=0,2,4,\cdots$ .

Similarly, the re-expansion of the odd series of eq14 using eq9 and eq13 gives

$\boldsymbol{\mathit{n}}$	$\boldsymbol{\mathit{\psi_n(y)}}$
$1$	$C_1ye^{-\frac{y^2}{2}}$
$3$	$C_1\biggr\[y-\frac{2(n-1)}{3!}y^3\biggr\]e^{-\frac{y^2}{2}}$
$5$	$C_1\biggr\[y-\frac{2(n-1)}{3!}y^3+\frac{2^2(n-1)(n-3)}{5!}y^5\biggr\]e^{-\frac{y^2}{2}}$
$7$	$C_1\biggr\[y-\frac{2(n-1)}{3!}y^3+\frac{2^2(n-1)(n-3)}{5!}y^5-\frac{2^3(n-1)(n-3)(n-5)}{7!}y^7\biggr\]e^{-\frac{y^2}{2}}$

Comparing the coefficients of each odd series in the above table, we have

$\psi_n(y)=C_1\biggr\[y-\frac{2(n-1)}{3!}y^3+\frac{2^2(n-1)(n-3)}{5!}y^5+\cdots+\frac{2^n$\frac{n-1}{2}$!}{(-1)^{\frac{n-1}{2}}2(n!)}y^n\biggr\]e^{-\frac{y^2}{2}}\;\;\;\;\;\;\;\;15b$

where $n=1,3,5,\cdots$ .

The recurrence relation of eq9 that led to eq14 only defines the value of $C_n$ in terms of $C_{n+2}$ . This implies that there are many solution sets depending on the value of $C_{n+2}$ . The convention is to set the leading coefficients (i.e. the coefficients of $y^n$ ) of eq15a and eq15b, which are independent of each other, as $2^n$ . Therefore,

$C_0\frac{2^n$\frac{n}{2}$!}{(-1)^{\frac{n}{2}}n!}=2^n\;\;or\;\;C_0=(-1)^{\frac{n}{2}}\frac{n!}{(\frac{n}{2})!}\;\;\;\;\;\;\;\;16$

and

$C_1\frac{2^n$\frac{n-1}{2}$!}{(-1)^{\frac{n-1}{2}}2(n!)}=2^n\;\;or\;\;C_1=(-1)^{\frac{n-1}{2}}\frac{2(n!)}{(\frac{n-1}{2})!}\;\;\;\;\;\;\;\;17$

Substituting eq16 and eq17 in eq14, we have $\psi(y)=H_n(y)e^{-\frac{y^2}{2}}$ , where $H_n(y)$ are the Hermite polynomials:

$H_n(y)=\biggr\{\begin{matrix}\biggr\{(-1)^{\frac{n}{2}}\frac{n!}{(\frac{n}{2})!}+\sum_{l=1}^{m}\biggr\[-\[2\epsilon-1-2(2l-2)\]\frac{C_{2l-2}}{\frac{(2l)!}{(2l-2)!}}\biggr\]y^{2l}\biggr\}e^{-\frac{y^2}{2}}&even\;series\;l=1,2,\cdots\\\biggr\{(-1)^{\frac{n-1}{2}}\frac{2(n!)}{(\frac{n-1}{2})!}y+\sum_{l=1}^{m}\biggr\[-\[2\epsilon-3-2(2l-2)\]\frac{C_{2l-1}}{\frac{(2l+1)!}{(2l-1)!}}\biggr\]y^{2l+1}\biggr\}e^{-\frac{y^2}{2}}&odd\;series\;l=1,2,\cdots\end{matrix}\;\;\;\;18$

The first few Hermite polynomials, un-normalised eigenfunctions of eq6 and their corresponding eigenvalues are

When $y=0$ , we can express the Hermite polynomials for odd $n$ as

$H_{2n+1}(0)=0\;\;\;n=0,1,2,\cdots\;\;\;\;\;\;\;\;19$

Similarly, when $y=0$ , we can express the Hermite polynomials for even $n$ as

$H_{2n}(0)=\frac{(-1)^n(2n)!}{n!}\;\;\;n=0,1,2,\cdots\;\;\;\;\;\;\;\;20$

Eq19 and eq20 are useful in proving the generating function for Hermite polynomials.

The table above also shows that the energy states $E_n$ of a quantum-mechanical harmonic oscillator are quantised:

$E_n=\epsilon\hbar\omega=\biggr$n+\frac{1}{2}\biggr$\hbar\omega\;\;\;n=0,1,2,\cdots\;\;\;\;\;21$

with the ground state $E_0=\frac{1}{2}\hbar\omega$ being non-zero. This energy is called the zero-point energy of the harmonic oscillator.

Question

Using the uncertainty principle, explain why the ground state of a quantum harmonic oscillator is non-zero.

Answer

If $E_0=0$ , then both the kinetic energy $E_k=\frac{p^2}{2m}$ and the potential energy $V$ of the oscillator are zero. $E_k=0$ implies that the momentum $p$ is zero and $\Delta p=0$ . $V=0$ means that the atom is at the equilibrium position, where $\Delta x=0$ . This would violate the uncertainty principle.

Since $H_n(y)$ is a solution to eq8, we can rewrite eq8 as

$\frac{d^2H_n(y)}{dy^2}-2y\frac{dH_n(y)}{dy}+(2\epsilon-1)H_n(y)=0\;\;\;\;\;\;\;\;22$

To complete the prove that $\psi_n(y)=H_n(y)e^{-\frac{y^2}{2}}$ is a solution to eq6, we will derive the normalisation constant $N_n$ of $\psi_n(y)=N_nH_n(y)e^{-\frac{y^2}{2}}$ and prove that the wavefunctions are orthogonal to one another in the next few articles.

Question

Do we have to change the variable $y$ back to $x$ ?

Answer

The Hermite polynomials are usually expressed in terms of $y$ because $H_n\biggr$\sqrt{\frac{m\omega}{\hbar}}x\biggr$$ looks more complicated:

Hermite polynomials

$H_0\biggr$\sqrt{\frac{m\omega}{\hbar}}x\biggr$=1$

$H_1\biggr$\sqrt{\frac{m\omega}{\hbar}}x\biggr$=2\sqrt{\frac{m\omega}{\hbar}}x$

$H_2\biggr$\sqrt{\frac{m\omega}{\hbar}}x\biggr$=4\biggr$\sqrt{\frac{m\omega}{\hbar}}x\biggr$^2-2$

$\vdots$
However, the normalisation constant is defined as an integral with respect to $x$ because $x$ is the variable representing position in the one-dimensional space that the oscillator is moving in.

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April 30, 2024

Orthogonality of the wavefunctions of the quantum harmonic oscillator

The orthogonality of the wavefunctions of the quantum harmonic oscillator $\psi_n(y)=H_ne^{-\frac{y^2}{2}}$ can be proven using the Hermite differential equation.

Substituting eq13 in eq22, we have

$\frac{d^2H_n}{dy^2}-2y\frac{dH_n}{dy}+2nH_n=0\;\;\;\;\;\;\;\;23$

where $H_n$ are the Hermite polynomials.

Question

Show that eq23 can be written as

$e^{y^2}\frac{d}{dy}\biggr$e^{-y^2}\frac{dH_n}{dy}\biggr$+2nH_n=0\;\;\;\;\;\;\;\;24$

Answer

Carrying out the derivative, the LHS of eq24 becomes

$e^{y^2}\biggr\[e^{-y^2}\frac{d^2H_n}{dy^2}+\frac{dH_n}{dy}(-2y)e^{-y^2}\biggr\]+2nH_n=\frac{d^2H_n}{dy^2}-2y\frac{dH_n}{dy}+2nH_n$

Multiplying eq24 (with a change of index from $n$ to $m$ ) by $H_n$ and subtracting the result from the product of $H_m$ and eq24, we have

${H_me^{y^2}\frac{d}{dy}\biggr$e^{-y^2}\frac{dH_n}{dy}\biggr$+2nH_mH_n-\biggr\[H_ne^{y^2}\frac{d}{dy}\biggr$e^{-y^2}\frac{dH_m}{dy}\biggr$+2mH_mH_n\biggr\]=0}$

$e^{-y^2}2(m-n)H_mH_n=H_m\frac{d}{dy}\biggr$e^{-y^2}\frac{dH_n}{dy}\biggr$-H_n\frac{d}{dy}\biggr$e^{-y^2}\frac{dH_m}{dy}\biggr$\;\;\;\;\;\;\;\;25$

Substituting ${H_m\frac{d}{dy}\biggr$e^{-y^2}\frac{dH_n}{dy}\biggr$=\frac{d}{dy}\biggr$H_me^{-y^2}\frac{dH_n}{dy}\biggr$-e^{-y^2}\frac{dH_n}{dy}\frac{dH_m}{dy}}$ and ${H_n\frac{d}{dy}\biggr$e^{-y^2}\frac{dH_m}{dy}\biggr$=\frac{d}{dy}\biggr$H_ne^{-y^2}\frac{dH_m}{dy}\biggr$-e^{-y^2}\frac{dH_m}{dy}\frac{dH_n}{dy}}$ in eq25 gives

$e^{-y^2}2(m-n)H_mH_n=\frac{d}{dy}\biggr\[e^{-y^2}\biggr$H_m\frac{dH_n}{dy}-H_n\frac{dH_m}{dy}\biggr$\biggr\]$

Integrating both sides of the above equation with respect to $y$ ,

$2(m-n)\int_{-\infty}^{\infty}e^{-y^2}H_mH_ndy=\int_{-\infty}^{\infty}\frac{d}{dy}\biggr\[e^{-y^2}\biggr$H_m\frac{dH_n}{dy}-H_n\frac{dH_m}{dy}\biggr$\biggr\]dy=0$

If $m\neq n$ , then $\int_{-\infty}^{\infty}e^{-y^2}H_mH_ndy=0$ . Since $\psi_n(y)=H_ne^{-\frac{y^2}{2}}=\psi^*_n(y)$ ,

$\int_{-\infty}^{\infty}\psi^*_m(y)\psi_n(y)dy=0\;\;\;\;m\neq n\;\;\;\;\;\;\;\;25a$

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