Advanced Chemistry - Mono Mole

June 28, 2025

Solid-liquid phase diagrams

Solid-liquid phase diagrams illustrate the relationship between temperature and composition in mixtures as they transition between solid and liquid states. These diagrams are essential for understanding melting behavior, phase equilibrium and solubility in binary or multi-component systems. They typically feature regions representing pure solid, pure liquid, and a mixture of both, separated by boundary lines such as the liquidus (liquid composition curve) and solidus (solid composition curve).

Ideal liquid solution and ideal solid solution

For instance, the silicon-germanium solid-liquid temperature-composition diagram resembles the liquid–vapour diagram of an ideal liquid solution (see above diagram). This similarity arises because the two elements form both an ideal solid solution and an ideal liquid solution. The region below the solidus corresponds to a single-phase solid solution, while the area above the liquidus represents a single-phase liquid solution. Between these two curves, the solid and liquid phases coexist in equilibrium, and the lever rule can be applied to determine the relative amounts of each phase at a given temperature.

Question

What is a solid solution?

Answer

A solid solution is a homogeneous mixture of two or more substances in the solid state, where one substance (the solute) is dissolved in another (the solvent) to form a single, uniform phase with a consistent crystal structure. Solute atoms either replace solvent atoms within the crystal lattice (substitutional solid solution) or occupy the spaces between them (interstitial solid solution). Essentially, it’s a “solid solution” of metals, often referred to as an alloy. For example, brass is a solid solution of copper (solvent) and zinc (solute).

Eutectics

The temperature-composition diagram of the gold-copper system is shown in the diagram below. In this system, gold and copper are completely miscible in both the liquid and solid phases, but they do not form an ideal solid solution due to differences in atomic interactions. The diagram features a eutectic point and resembles a minimum azeotrope in shape. The eutectic point corresponds to the lowest melting temperature of the alloy, occurring at a specific composition known as the eutectic composition. At this point, the homogeneous liquid solidifies into a single solid phase with a well-defined composition. Because the eutectic temperature is lower than the melting points of either pure component, this point is particularly useful in applications such as casting and welding. The term “eutectic” was coined in 1884 by British physicist and chemist Frederick Guthrie, derived from the Greek words eu (“well” or “good”) and têxis (“melting”), meaning “easily melted.”

Unlike the gold-copper system, the silver-copper system features complete miscibility of its components in the liquid phase but only partial miscibility in the solid phase. Based on the gold-copper phase diagram, we would expect the solid solution region to divide into multiple regions due to the limited solid-state solubility of silver and copper at lower temperatures (see diagram below).

Indeed, the temperature-composition diagram of the silver-copper system includes a single-phase liquid solution region ( $ls$ ), two single-phase solid solution regions ( $ss_{\alpha}$ and $ss_{\beta}$ ), a two-phase solid region ( $\alpha+\beta$ ), and two two-phase regions where a liquid solution is in equilibrium with a solid solution ( $ls+ss_{\alpha}$ and $ls+ss_{\beta}$ ). The eutectic point k, at 779^oC, represents the equilibrium state where the solid solutions $ss_{\alpha}$ and $ss_{\beta}$ coexist with the liquid phase .

When a liquid solution with composition at point d is cooled, the solid solution phase $ss_{\alpha}$ begins to precipitate at point e, with its composition corresponding to point h on the tie-line. As the two-phase mixtrure of $ls+ss_{\alpha}$ cools to point f, it reaches the eutectic composition. At this stage, the solid solution $ss_{\beta}$ also begins to form.

The Ag–Cu phase diagram is a powerful tool for understanding, designing and processing silver–copper alloys. It helps engineers to select the right alloy composition for specific properties (e.g. ductility and conductivity), and guide thermal treatments for practical applications in electronics, jewelry and joining technologies.

Compound formation (congruent melting)

Consider two substances, A and B, that react to produce a compound C in a 1:1 ratio. If C separately forms eutectic mixtures with both A and B, the solid-liquid phase diagram of A and B (e.g. aniline and phenol) will appear as follows:

Aniline $C_6H_5NH_2$ and phenol $C_6H_5OH$ combine to form an adduct $C_6H_5OH\cdot C_6H_5NH_2$ , which is a stable complex with its own characteristic melting point and crystal structure distinct from the individual components. Aniline and phenol are miscible in the liquid phase at higher temperatures, while the adduct does not exist in the liquid state. Furthermore, the solid forms of all three species are completely insoluble in one another. As a result, solid phenol coexists with the solid adduct in a two-phase region between $0\leq x_{aniline}\leq 0.5$ at lower temperatures. When $x_{aniline}\geq 0.5$ , all phenol molecules will react with aniline, leaving a two-phase region consisting of solid adduct and solid aniline. In other words, the maximum amount of adduct is formed when $x_{aniline}=0.5$ . Since the adduct forms eutectic mixtures with both aniline and phenol, the phase diagram can be viewed as two eutectic phase diagrams positioned side by side. Points a and b correspond to the eutectic points of the phenol-adduct and the adduct-aniline mixtures respectively.

Consider the isopleth defg. As the mixture cools from point e to point g, negligible amounts of phenol will be present, meaning that the tie-lines extend only from $x_{aniline}=0.5$ onwards. On the other hand, if a solution with $x_{aniline}=0.5$ is cooled, only pure solid adduct will separate out, without any change in composition. If the process is reversed, the liquid formed will have the same composition as the solid. This phenomenon is known as congruent melting.

In pharmaceutical and chemical industries, the diagram can help in purification processes or the synthesis of the adduct of aniline and phenol, which is a valuable intermediate in the production of various compounds, such as dyes, drugs, and plastics

Compound formation (incongruent melting)

Some stable solids of the form A₂B, each melts to give a liquid with a different composition. We call such a process incongruent melting. The phase diagram of an A₂B alloy resembles that of the aniline-phenol system, except that the melting point of one component is much higher the other (see diagram below).

The vertical line at $x_{A}=0.67$ in the phase diagram marks the composition of the pure solid compound A₂B. When the alloy at $x_{A}=0.67$ is heated, it begins to melt at temperature $T_{1}$ , producing pure solid A and a single-phase liquid containing both A and B. The resulting liquid has a mole fraction of A ( $x_{A'}$ ) that differs from that of A₂B.

An example of such a system is the Na₂K alloy, which is used in liquid metal coolants and metallic lubricants. Understanding the melting behavior of Na₂K is important for selecting suitable compositions in these applications. Although the phase diagram shows the presence of Na₂K, it is still a binary system diagram for the Na-K system. The compound Na₂K is simply an intermetallic phase that appears as part of the equilibrium between Na and K. In contrast, ternary phase diagrams involve three independent components.

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June 28, 2025

Ternary systems

A ternary system refers to a mixture composed of three components. Its phase diagram requires at least three dimensions, typically representing either temperature $T$ or pressure $p$ , along with two independent mole fractions.

Question

Show that the height $h$ of an equilateral triangle is $h=\frac{\sqrt{3}}{2}s$ , where $s$ is the side length, and hence, $DE+DF+DG=h$ .

Answer

Using Pythagorean theorem, $\biggr$\frac{s}{2}\biggr$^2+h^2=s^2$ , which rearranges to $h=\frac{\sqrt{3}}{2}s$ . The area of the triangle is $\frac{1}{2}sh=\frac{\sqrt{3}}{4}s^2$ . Since $\Delta_{BCD}+\Delta_{BAD}+\Delta_{ACD}=\frac{\sqrt{3}}{4}s^2$ , we have $\frac{s}{2}(DE+DF+DG)=\frac{\sqrt{3}}{4}s^2$ , which rearranges to $DE+DF+DG=h$ .

Since $DE+DF+DG=h$ , the perpendicular distances DE, DF and DG becomes $x_C$ , $x_B$ and $x_A$ respectively if we set $h=1$ . This normalisation allows us to represent the ternary system in an equilateral triangle, assuming both $T$ and $p$ are held constant (see diagram above). For instance, the red dot in the diagram corresponds to mole fractions $x_A=0.4$ , $x_B=0.2$ and $x_C=0.4$ . In fact, once two independent mole fractions are specified, the third is automatically determined due to the constraint $x_A+x_B+x_C=1$ . This effectively reduces the degrees of freedom to two, allowing the system to be represented on a two-dimensional diagram.

The ternary phase diagram below represents the acetone-water-diethyl ether system at 30^oC and 1 atm. Under these conditions, water and diethyl ether are only partially miscible, while the other two binary pairs are fully miscible. The region under the curve consists of two liquid phases in equilibrium, whereas the region above the curve represents a single-phase liquid.

Since temperature and pressure are constant, tie lines must remain in the plane of the diagram, but they are not required to be parallel. Their orientations are determined experimentally. The compositions of the two coexisting phases at a given point under the curve are found at the ends of the tie line that intersects the point. For example, point f lies within the two-phase region and corresponds to a water-rich, ether-poor phase ( $\alpha$ ) of composition e, and a water-poor, ether-rich phase ( $\beta$ ) of composition g. Moreover, point e is richer in acetone than point g. Finally, the lever rule also applies here: the relative amounts of the two phases are inversely proportional to the lengths of the tie line segments, i.e. $l_{ef}n_{\alpha}=l_{fg}n_{\beta}$ .

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June 28, 2025

Phase rule

The phase rule, formulated by Josiah Willard Gibbs, describes the number of degrees of freedom (independent variables) needed to define a multiphase, multicomponent system at equilibrium. It is given by

$F=C-P+2\;\;\;\;\;\;\;\;203$

where

$F$ is the number of degrees of freedom (i.e. independent intensive variables such as temperature, pressure or mole fraction). This value is a non-negative integer.
$C$ is the number of independent components that can describe the composition of a point in a phase diagram.
$P$ is the number of phases present in equilibrium.

For example, we need two independent mole fractions to describe a liquid mixture of three fully miscible components at constant temperature and pressure (single-phase ternary system) because $C=3$ and $P=1$ .

To derive eq203, assume that every chemical species is present in every phase of the system. Each phase has a composition defined by the mole fractions of $C$ components. Since mole fractions must sum to 1, each phase has $C-1$ independent composition variables. For $P$ phases, the total number of composition variables is $P(C-1)$ . Including temperature and pressure, the total number of variables is $P(C-1)+2$ .

However, not all of these variables are independent because the system is at equilibrium. At equilibrium, the chemical potential of each component must be equal in all phases. For example, for component 1,

$\mu_1^{\alpha}=\mu_1^{\beta}=\cdots=\mu_1^{p}$

This gives $P-1$ equations per component, resulting in a total of $(P-1)C$ independent equations. These equations represent constaints that reduce the number of independent variables because each chemical potential is a function of temperature, pressure and the mole fractions of all components in each phase, e.g. $\mu_1^{\alpha}=\mu_1^{\alpha}(T,p,x_1^{\alpha},x_2^{\alpha},\cdots,x_C^{\alpha})$ . If say, $\mu_1^{\alpha}=AT+Bp+Cx_1^{\alpha}+\cdots$ and $\mu_1^{\beta}=ET+Fp+Gx_1^{\beta}+\cdots$ , then one variable becomes dependent on the others when we equate $\mu_1^{\alpha}=\mu_1^{\beta}$ . Even if the functions are more complex, the value of any one variable remains dependent on the others.

Therefore, the number of independent variables needed to define a multiphase, multicomponent system at equilibrium is $F=P(C-1)+2-(P-1)C$ , which rearranges to eq203. In this derivation, we assumed that every chemical species is present in every phase of the system. What if one or more species is absent from one or more phases? Suppose species $i$ is absent from phase $\delta$ . In this case, the number of variables is reduced by one because $x_i^{\delta}$ is no longer a variable. At the same time, the number of independent chemical potential equations is also reduced by 1, e.g. $\mu_i^{\alpha}=\mu_i^{\beta}=\mu_i^{\delta}$ becomes $\mu_i^{\alpha}=\mu_i^{\beta}$ . Therefore, the phase rule still holds.

Question

Apply the phase rule to 1) a system where calcium carbonate, calcium oxide and carbon dioxide are at equilibrium, 2) points A to F indicated in the one-component phase diagram above, and explain why four phases cannot mutually coexist in equilibrium for this case.

Answer

There are clearly three phases ( $P=3$ ): $CaCO_3(s)$ , $CaO(s)$ and $CO_2(g)$ . Although there are three chemical species in the system, the reaction $CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)$ introduces a constraint, reducing the number of independent components to two ( $C=2$ ). Therefore, $F=1$ . This implies that if one variable is fixed, the other is automatically determined. For instance, at a given temperature, there’s a unique equilibrium pressure of $CO_2$ where all three phases can coexist.

Point	C	P	F
A	1	1	2
B	1	2	1
C	1	3	0
D	1	1	2
E	1	1	2
F	1	2	1

When only one phase is present in a one-component system (points A, D and E), pressure and temperature can be varied independently within that single-phase region, giving two degrees of freedom. Points on phase boundaries (such as B and F) represent the coexistence of two phases in equilibrium. These points have only one degree of freedom, meaning that if pressure changes, temperature must adjust accordingly to stay on the phase boundary (and vice versa). Point C, the triple point, represents the unique conditions where three phases coexist in equilibrium. For a one-component system, this corresponds to zero degrees of freedom as temperature and pressure are fixed. If we consider four phases ( $P=4$ ) in a one-component system ( $C=1$ ), we find $F=-1$ , which is not physically meaningful. This confirms that no more than three phases can coexist in equilibrium in a one-component system.

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December 29, 2024February 8, 2025

Why does a current flowing through a wire generate a magnetic field?

Why does a current flowing through a wire generate a magnetic field? The answer lies in special relativity.

Consider a current flowing through a straight length of wire. In the rest frame of the wire, the positive ions (nuclei) are stationary, while the electrons are moving. Suppose a negative test charge is located at a distance from the wire, and moves parallel to the wire with the same velocity as the electrons. In this frame, the wire is electrically neutral overall because the number of positive charges equals the number of negative charges. Consequently, there is no net static electric field acting on the test charge in this frame. However, we observe that the test charge moves in response to a force towards the wire.

In the frame of the moving test charge, the test charge and the electrons are stationary, while the positive ions move in the opposite direction. Due to length contraction, the distance between the moving positive ions appears shorter to the test charge, leading to an increased positive charge density in the test charge’s frame. As a result, the wire has an apparent higher positive charge density compared to the negative charge density of the stationary electrons. This creates a net electric field, which exerts a force on the test charge towards the wire. In the rest frame of the wire, this force is referred to as a magnetic force. Therefore, both frames yield the same total force on the test charge, but it is described as a magnetic force in the wire’s rest frame and an electric force in the test charge’s frame. Since we can mathematically describe the magnetic force in the wire’s rest frame by defining a magnetic field, we say that a current flowing in a wire generates a magnetic field.

In summary, the electric field and the magnetic field are simply different observations of the same underlying phenomenon. Just as a coin has two faces that are inseparable yet different, electric and magnetic fields are inseparable parts of the same phenomenon, with their roles depending on the relative motion of the observer.

Question

Doesn’t length contraction also apply to the moving electrons in the wire’s rest frame? If so, why is there no net electric field acting on the test charge in that frame?

Answer

The rest frame of the wire is the initial premise. In this frame, the wire is observed to be electrically neutral, with length contraction already accounted for in this electrical neutrality. The question then shifts to how the situation differs in another reference frame, based on the initial premise.

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September 23, 2024

Laguerre polynomials

Laguerre polynomials $v(x)$ are a sequence of polynomials that are solutions to the Laguerre differential equation:

$x\frac{d^2v}{dx^2}+(1-x)\frac{dv}{dx}+nv=0\;\;\;\;\;\;\;\;420$

where $n$ is a constant.

When $x=0$ , eq420 simplifies to $\frac{dv}{dx}+nv=0$ . The solution to this first-order differential equation is $v(x)=Ae^{-nx}$ , which can be expressed as the Taylor series $v(x)=A\sum_{n=0}^{\infty}\frac{(-nx)^n}{n!}$ . This implies that eq420 has a power series solution around $x=0$ . To determine the exact form of the power series solution to eq420, let $v(x)=\sum_{j=0}^{\infty}a_jx^j$ .

Substituting $v$ , $v'$ and $v''$ in eq420 yields

$x\sum_{j=0}^{\infty}j(j-1)a_jx^{j-2}+(1-x)\sum_{j=0}^{\infty}ja_jx^{j-1}+n\sum_{j=0}^{\infty}a_jx^{j}=0\;\;\;\;\;\;\;\;421$

$\sum_{j=0}^{\infty}j^2a_jx^{j-1}+\sum_{j=0}^{\infty}(n-j)a_jx^{j}=0$

Setting $j=j+1$ in the first sum,

$\sum_{j=-1}^{\infty}(j+1)^2a_{j+1}x^{j}+\sum_{j=0}^{\infty}(n-j)a_jx^{j}=\sum_{j=0}^{\infty}(j+1)^2a_{j+1}x^{j}+\sum_{j=0}^{\infty}(n-j)a_jx^{j}=0$

$\sum_{j=0}^{\infty}\biggr\[(j+1)^2a_{j+1}+(n-j)a_j\biggr\]x^{j}=0\;\;\;\;\;\;\;\;422$

Eq422 is only true if all coefficients of $x$ in is 0 (see this article for explanation). So, $(j+1)^2a_{j+1}+(n-j)a_j=0$ , or equivalently,

$a_{j+1}=\frac{j-n}{(j+1)^2}a_j\;\;\;\;\;\;\;\;423$

Eq423 is a recurrence relation. If we know the value of $a_0$ , we can use the relation to find $a_1,a_2,...$ .

$\boldsymbol{\mathit{j}}$	Recurrence relation
$0$	$a_1=-na_0$
$1$	$a_2=\frac{1-n}{(1+1)^2}a_1=\frac{n(n-1)}{1^22^2}a_0$
$2$	$a_3=\frac{2-n}{(2+1)^2}a_2=-\frac{n(n-1)(n-2)}{1^22^23^2}a_0$
$\vdots$	$\vdots$

Comparing the recurrence relations, we have

$a_k=(-1)^k\frac{n(n-1)(n-2)\cdots (n-k+1)}{1^22^2\cdots k^2}a_0$

$a_k=(-1)^k\frac{n!}{(k!)^2(n-k)!}\;\;\;\;\;\;\;\;424$

where $a_0=1$ by convention (so that $L_n(0)=1$ ).

Letting $k=j$ in eq424 and substituting it in $v(x)=\sum_{j=0}^{n}a_jx^j$ yields the Laguerre polynomials:

$L_n(x)=\sum_{j=0}^{n}(-1)^j\frac{n!}{(j!)^2(n-j)!}x^j\;\;\;\;\;\;\;\;425$

where we have replaced $v(x)$ with $L_n(x)$ .

The first few Laguerre polynomials are:

$L_0(x)=1$

$L_1(x)=-x+1$

$L_2(x)=\frac{1}{2}(x^2-4x+2)$

$L_3(x)=\frac{1}{6}(-x^3+9x^2-18x+6)$

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September 23, 2024

Rodrigues’ formula for the Laguerre polynomials

The Rodrigues’ formula for the Laguerre polynomials is a mathematical expression that provides a method to calculate any Laguerre polynomial using differentiation.

It is given by

$L_n(x)=\frac{e^x}{n!}\frac{d^n}{dx^n}(x^ne^{-x})\;\;\;\;\;\;\;\;428$

To prove eq428, we apply Leibniz’ theorem as follows:

$L_n(x)=\frac{e^x}{n!}\sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}\frac{d^kx^n}{dx^k}\frac{d^{n-k}e^{-x}}{dx^{n-k}}$

Substituting $\frac{d^kx^n}{dx^k}=\frac{n!}{(n-k)!}x^{n-k}$ and $\frac{d^{n-k}e^{-x}}{dx^{n-k}}=(-1)^{n-k}e^{-x}$ in the above equation and rearranging yields

$L_n(x)=\sum_{k=0}^n(-1)^{n-k}\frac{n!}{k!(n-k)!(n-k)!}x^{n-k}$

Letting $j=n-k$ , we have $L_n(x)=\sum_{j=0}^{n}(-1)^j\frac{n!}{(j!)^2(n-j)!}x^j$ , which is the expression for the Laguerre polynomials

Question

How do we change the variable in the summation $\sum_{k=0}^{n}$ by letting $j=n-k$ ?

Answer

When $k=0$ , $j=n$ , and when $k=n$ , $j=0$ . So, $\sum_{k=0}^{n}=\sum_{j=n}^{0}$ . Reversing the summation order, $\sum_{k=0}^{n}=\sum_{j=0}^{n}$ .

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September 23, 2024May 8, 2025

Generating function for the Laguerre polynomials

The generating function $G(x,t)$ for the Laguerre polynomials is a mathematical tool that, when expanded as a power series, produces Laguerre polynomials as its coefficients in terms of a variable.

$G(x,t)=\frac{e^{-\frac{xt}{1-t}}}{1-t}=\sum_{n=0}^{\infty}L_n(x)t^n\;\;\;\;\;\;\;\;430$

Eq430 means that the cofficient of $t^n$ in the expansion of $\frac{e^{-\frac{xt}{1-t}}}{1-t}$ is $L_n(x)$ . To prove this, we expand the exponential term as a Taylor series:

$\frac{e^{-\frac{xt}{1-t}}}{1-t}=\frac{1}{1-t}\sum_{n=0}^{\infty}\frac{$-\frac{xt}{1-t}$^n}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^nt^n(1-t)^{-(n+1)}$

Expanding $(1-t)^{-(n+1)}$ as a binomial series gives

$\frac{e^{-\frac{xt}{1-t}}}{1-t}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^nt^n\sum_{k=0}^{\infty}\begin{pmatrix}-n-1\\k\end{pmatrix}(-t)^{k}$

Since

$\begin{align}\begin{pmatrix}-n-1\\k\end{pmatrix}&=\frac{(-n-1)!}{k!(-n-k-1)!}=\frac{(-n-1)(-n-2)\cdots (-n-k)}{k!}\\&=\frac{(-1)^k(n+1)(n+2)\cdots(n+k)}{k!}=\frac{(-1)^k(n+k)!}{n!k!}\end{align}$

we have

$\frac{e^{-\frac{xt}{1-t}}}{1-t}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^nt^n\sum_{k=0}^{\infty}\frac{(n+k)!}{n!k!}t^{k}=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}(-1)^n\frac{(n+k)!}{k!(n!)^2}x^nt^{n+k}$

Letting $n+k=m$

$\frac{e^{-\frac{xt}{1-t}}}{1-t}=\sum_{n=0}^{\infty}\sum_{m=n}^{\infty}(-1)^n\frac{m!}{(m-n)!(n!)^2}x^nt^{m}\;\;\;\;\;\;\;\;431$

We now have a sum over $m$ and then over $n$ . Since $m=n+k$ and both $n$ and $k$ range from $0$ to $\infty$ , the sum over $m$ ranges from $0$ to $\infty$ . The new range of $n$ in the outer sum is determined by the conditions that $n\geq 0$ and $n=m-k$ , where $m$ and $k$ range from $0$ to $\infty$ . Consequently, $n$ has a lower limit of 0 and an upper limit of $m$ . Eq431, after swapping the order of summation, then becomes

$\frac{e^{-\frac{xt}{1-t}}}{1-t}=\sum_{m=0}^{\infty}\biggr\[\sum_{n=0}^{m}(-1)^n\frac{m!}{(m-n)!(n!)^2}x^n\biggr\]t^{m}\;\;\;\;\;\;\;\;431$

where $\sum_{n=0}^{m}(-1)^n\frac{m!}{(m-n)!(n!)^2}x^n$ are the Laguerre polynomials.

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September 23, 2024

Recurrence relations of the Laguerre Polynomials

The recurrence relations of the Laguerre polynomials describe how each polynomial in the sequence can be obtained from its predecessors.

Some useful recurrence relations of the Laguerre polynomials include

$(n+1)L_{n+1}=(2n+1-x)L_n-nL_{n-1}\;\;\;\;\;\;\;\;434$

$L_{n-1}=L_{n-1}'-L_{n}'\;\;\;\;\;\;\;\;435$

$xL_{n}'=nL_{n}-nL_{n-1}\;\;\;\;\;\;\;\;436$

To derive eq434, differentiate eq430 with respect to $t$ to give

$\frac{\partial G}{\partial t}=\frac{1-x-t}{(1-t)^3}e^{-\frac{xt}{1-t}}=\sum_{n=0}^{\infty}nL_nt^{n-1}$

Substituting eq430 gives

$(1-x-t)\sum_{n=0}^{\infty}L_nt^{n}=(1-t)^2\sum_{n=0}^{\infty}nL_nt^{n-1}$

Equating the coefficients of $t^n$ yields

$(1-x)L_{n}-L_{n-1}=(n+1)L_{n+1}-2nL_{n}+(n-1)L_{n-1}$

which rearranges to eq434.

To derive eq435, differentiate eq430 with respect to $x$ to give

$\frac{\partial G}{\partial t}=-\frac{t}{(1-t)^2}e^{-\frac{xt}{1-t}}=\sum_{n=0}^{\infty}L_n't^{n}$

Substituting eq430 gives

$-t\sum_{n=0}^{\infty}L_nt^{n}=(1-t)\sum_{n=0}^{\infty}L_n't^n$

Equating the coefficients of $t^n$ yields eq435. To derive eq436, differentiate eq434 with respect to $x$ to yield

$(n+1)L_{n+1}'=(2n+1-x)L_n'-L_n-nL_{n-1}'$

Substituting eq435 gives

$(n+1)L_{n+1}'=(2n+1-x)L_n'-L_n-n(L_{n-1}+L_n')$

Letting $n=n+1$ in eq435, substituting the result in the above equation and rearranging yields eq436.

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September 23, 2024

Associated Laguerre polynomials

The associated Laguerre polynomials are a sequence of polynomials that are solutions to the associated Laguerre differential equation:

$x\frac{d^2L_n^m(x)}{dx^2}+(m+1-x)\frac{dL_n^m(x)}{dx}+nL_n^m(x)=0\;\;\;\;\;\;\;442$

where $L_n^m(x)=\frac{d^m}{dx^m}L_{n+m}(x)$ are the associated Laguerre polynomials.

To show that $L_n^m$ are solutions to eq442, we refer to eq420, where $x\frac{d^2L_n}{dx^2}+(1-x)\frac{dL_n}{dx}+nL_n=0$ . Letting $n=n+m$ , we have $x\frac{d^2L_{n+m}}{dx^2}+(1-x)\frac{dL_{n+m}}{dx}+(n+m)L_{n+m}=0$ . Differentiating this equation $m$ times with respect to $x$ gives

$\frac{d^m}{dx^m}\biggr$x\frac{d^2L_{n+m}}{dx^2}\biggr$+\frac{d^m}{dx^m}\biggr\[(1-x)\frac{dL_{n+m}}{dx}\biggr\]+(n+m)\frac{d^mL_{n+m}}{dx^m}=0$

Applying Leibniz’ theorem,

$\sum_{k=0}^m\begin{pmatrix}m\\k\end{pmatrix}\frac{d^kx}{dx^k}\frac{d^{m-k+2}L_{n+m}}{dx^{m-k+2}}+\sum_{k=0}^m\begin{pmatrix}m\\k\end{pmatrix}\frac{d^k(1-x)}{dx^k}\frac{d^{m-k+1}L_{n+m}}{dx^{m-k+1}}+(n+m)\frac{d^mL_{n+m}}{dx^m}=0$

$x\frac{d^{m+2}L_{n+m}}{dx^{m+2}}+m\frac{d^{m+1}L_{n+m}}{dx^{m+1}}+(1-x)\frac{d^{m+1}L_{n+m}}{dx^{m+1}}-m\frac{d^{m}L_{n+m}}{dx^{m}}+(n+m)\frac{d^{m}L_{n+m}}{dx^{m}}=0$

$x\frac{d^{m+2}L_{n+m}}{dx^{m+2}}+(m+1-x)\frac{d^{m+1}L_{n+m}}{dx^{m+1}}+n\frac{d^{m}L_{n+m}}{dx^{m}}=0$

This implies that $L_n^m(x)=\frac{d^m}{dx^m}L_{n+m}(x)$ . Since $(-1)^m\frac{d^m}{dx^m}L_{n+m}(x)$ is also a solution to the associated Laguerre differential equation, $L_n^m(x)$ can also be expressed as

$L_n^m(x)=(-1)^m\frac{d^m}{dx^m}L_{n+m}(x)\;\;\;\;\;\;\;\;443$

When $m=0$ , eq442 becomes the Laguerre differential equation. Therefore, $L_n^0(x)=L_n(x)$ . Substituting eq425 in eq443 yields

$L_n^m(x)=(-1)^m\frac{d^m}{dx^m}\sum_{j=0}^{n+m}(-1)^j\frac{(n+m)!}{(j!)^2(n+m-j)!}x^j$

For $j<m$ , the terms in the summation equal zero. For $j\geq m$ , we note that $\frac{d^m}{dx^m}x^j=j(j-1)\cdots (j-m+1)x^{j-m}=\frac{j!}{(j-m)!}x^{j-m}$ , and so,

$L_n^m(x)=(-1)^m\sum_{j=m}^{n+m}(-1)^j\frac{(n+m)!}{j!(n+m-j)!(j-m)!}x^{j-m}$

Letting $k=j-m$ ,

$L_n^m(x)=\sum_{k=0}^{n}(-1)^k\frac{(n+m)!}{(k+m)!(n-k)!k!}x^{k}\;\;\;\;\;\;\;\;444$

Eq444 is the general expression for the un-normalised associated Laguerre polynomials. Using eq444, the first few associated Laguerre polynomials in terms of $m$ are

$L_0^m(x)=1$

$L_1^m(x)=1+m-x$

$L_2^m(x)=\frac{1}{2}[x^2-2(m+2)x+(m+1)(m+2)]$

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September 23, 2024

Orthogonality of the Laguerre polynomials

The orthogonality of the Laguerre polynomials states that the integral of the product of two distinct Laguerre polynomials over a specified interval is zero.

It is defined mathematically as:

$\int_0^{\infty}L_n(x)L_m(x)e^{-x}dx=0\;\;\;\;n\neq m\;\;\;\;\;\;\;438$

where $e^{-x}$ is known as a weight function.

Question

Why is the weight function included? Can the orthogonality of the Laguerre polynomials be defined as $\int_0^{\infty}L_n(x)L_m(x)dx=0$ , where $n\neq m$ ? Why are the limits of integration from 0 to $\infty$ ?

Answer

The weight function is an integral part of the orthogonality definition of Laguerre polynomials due to its role in ensuring convergence and its practical applications. It is often tied to specific problems, such as those in quantum mechanics. Omitting the weight function would sever this connection and could potentially alter the orthogonality properties of the polynomials. Therefore, defining orthogonality without the weight function would generally be invalid and would not reflect the intended use and properties of the Laguerre polynomials.

The weight function naturally defines the integration range because $e^{-x}\rightarrow 0$ as $x\rightarrow\infty$ , making the integral convergent over this range. This range is also connected to specific problems, such as the radial part of the wave functions in quantum mechanics. In the context of the hydrogen atom, $x$ represents a distance, which is always non-negative.

To prove eq438, we multiply eq430 for $L_n(x)$ and $L_m(x)$ to give

$\frac{e^{-\frac{xt}{1-t}-\frac{xs}{1-s}}}{(1-t)(1-s)}=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}L_n(x)L_m(x)t^ns^m$

Multiplying through by $e^{-x}$ and integrating with respect to $x$ yields

$\int_0^{\infty}\frac{e^{-(\frac{t}{1-t}+\frac{s}{1-s}+1)x}}{(1-t)(1-s)}dx=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\biggr\{\int_0^{\infty}e^{-x}L_n(x)L_m(x)dx\biggr\}t^ns^m$

which simplifies to

$\frac{1}{1-st}=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\biggr\{\int_0^{\infty}e^{-x}L_n(x)L_m(x)dx\biggr\}t^ns^m$

Expressing the LHS as a binomial series gives

$\sum_{n=0}^{\infty}(st)^n=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\biggr\{\int_0^{\infty}e^{-x}L_n(x)L_m(x)dx\biggr\}t^ns^m$

Equating the coefficients of $t^ns^m$ when $m\neq n$ gives eq438. If we further equate the coefficients of $t^ns^m$ when $m=n$ , we have $\int_0^{\infty}L_n(x)L_m(x)e^{-x}dx=1$ .

Therefore,

$\int_0^{\infty}L_n(x)L_m(x)e^{-x}dx=\delta_{mn}\;\;\;\;\;\;\;\;439$

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