The real hydrogen atom

6569 /www/nrich/html/content/id/6569/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><ul id="stemLinks"> <li><a href="http://nrich.maths.org/6558">Warm-up</a></li> <li><a href="http://nrich.maths.org/5937">Try this next</a></li> <li><a href="http://plus.maths.org/content/quantum-geometry">Read: mathematics</a></li> <li><a href="http://plus.maths.org/content/quantum-uncertainty">Read: science</a></li> <li><a href="http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydsch.html#c2">Explore further</a></li> </ul> <div> </div> The structures of atoms are described by quantum mechanics. This is a complicated theory which developed over time following a great deal of experimental input. Now, it is used routinely to model the behaviour of electrons, atoms and molecules. At its heart lies the Schrodinger equation, which is an equation for an object called the wavefunction, written as $\psi({\bf x}, t)$. The square of the wavefunction is a probability density function such that $|(\psi{\bf x}, t)|^2dV$ is the probability of finding the particle in a small ball of size $dV$ centred on ${\bf x}$ at time $t$. Solving the Schrodinger equation allows chemists to determine the physical form of the electron orbitals of atoms. The general Schrodinger equation is very complicated, but for the motion of an electron around a single hydrogen nucleus reduces to three relatively simple equations when we use spherical polar coordinates $(r, \theta, \phi)$ (in terms of the globe $\phi$ is the longitude and $\theta$ is the latitude of a point) to describe the location of the electron through three functions $R(r), P(\theta), F(\phi)$ $$ \frac{1}{R}\frac{d}{dr}\left[r^2 \frac{dR}{dr}\right] -\frac{8\pi^2\mu}{h^2}\left(Er^2+\alpha r\right)=l(l+1) $$ $$ \frac{\sin\theta}{P}\frac{d}{d\theta}\left[\sin\theta \frac{dP}{d\theta}\right] +l(l+1)\sin^2\theta =m^2_l $$ $$ \frac{1}{F}\frac{d^2F}{d\phi^2}=-m^2_l $$ These equations involve the effective mass $\mu$ of the electron in terms of the mass of the electron and proton $m_e$ and $m_p$ as $\mu = \frac{m_em_p}{m_e+m_p}$, $\alpha = \frac{e^2}{4\pi\epsilon_0}$, the energy $E$ of the particle and Planck's constant $h =6.626068\times 10^{-34}$ m$^2$ kg s$^{-1}$. There are three quantum numbers associated with these equations: The principal quantum number $n=1, 2, 3, \dots$ The orbital quantum numbers $l = 0, 1, \dots, n-1$. The magnetic quantum number $m= -l, -l+1, \dots, l -1, l$ The energy $E$ of the electron depends on the principal quantum number as $$ E=-\frac{\mu e^4}{8 n^2h^2\epsilon^2_0} $$ Solving this tricky set of equations gives the wave function for the electron to be $\psi({\bf x}, t)=R(r)P(\theta)F(\phi)$, from which the likely positions of the electron can be determined. The s, p, d and f shells correspond to solutions with $l=0,1,2$ and $3$ respectively. The goal of the quantum chemist is to attempt to extract solutions from these equations. Task 0: Familiarise yourself with the structure of these equations. For example, what does the lowest energy ($n=1$) set of equations look like? Task 1: Show that if $(R_1, P_1, F_1)$ is a solution to the set of equations then $(aR_1, bP_1, cF_1)$ will also be a solution for any non-zero constants $a, b$ and $c$. Task 2: Start simply: do constant values for either $R, P$ or $F$ satisfy the equations for particular choices of $n, l, m$? Task 3: Solving the $P$ equation will clearly be likely to involve trigonometric functions. Do $P=\sin\theta, \cos\theta, \tan\theta$ solve the $P$ equation for particular choices of $l$ and $m$? How about other simple combinations of $\sin$ and $\cos$? Task 4: The $R$ equation contains many constants. Group these together as $a_0 = \frac{\pi e^2\mu}{h^2\epsilon_0}$ and see how the equation simplifies a little. Task 5: On physical grounds we would expect the chance of the electron being very far away from the nucleus to be very small, tending to zero as $r$ increases. Why would a function $R = f(r)e^{-ar}$ satisfy this condition for any polynomial $f(r)$ and positive constant $a$? Try the simplest case $R=e^{-ar}$. Does this result in a solution? What about the possibilities for $R= (a+br)e^{-cr}$? for constants $a, b, c$? <div class="framework">NOTES AND BACKGROUND In this problem, we have only looked briefly into attempting to partially solve the Schrodinger equation for the simplest atom, and have taken the Schrodinger equation on trust. Exposure to the ideas raised in this problem will both improve your skills with differential equations and also ease the way into the true theory underlying quantum mechanics and atoms.To derive and solve these equations more generally requires some heavyweight analytical tools of the kind met at university. More complicated versions must be solved using clever numerical integration schemes. If you wish to find out more about these equations see, for example, <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydsch.html#c2">http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydsch.html#c2</a> </div></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">This problem is really pushing well into university concepts, but the solution given here is well worth a read by keen school students who wish to further their understanding of differential equations. Task 0: The lowest energy set of equation corresponds to $ n = 0$. This implies that $l =0$ and $m_l = 0$ also. Thus: a) $\frac{1}{R} \frac{d}{dr}\left[r^2 \frac{dR}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$ b) $\frac{sin\theta}{P} \frac{d}{d\theta} \left[sin\theta \frac{dP}{d\theta}\right] = 0$ c) $\frac{1}{F} \frac{d^2F}{d\phi^2} = 0$ Task 1: If $R_1$ is a solution, then for $aR_1$: $\frac{1}{aR_1} \frac{d}{dr}\left[r^2 \frac{d(aR_1)}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$ Since $a$ is a constant, it can be removed outside of the differential: $\frac{a}{aR_1} \frac{d}{dr}\left[r^2 \frac{dR_1}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$ $\frac{1}{R_1} \frac{d}{dr}\left[r^2 \frac{dR_1}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$ Therefore, since $R_1$ is a solution, so is $aR_1$ for any non-zero constant $a$. If $P_1$ is a solution, then for $bP_1$: $\frac{sin\theta}{bP_1} \frac{d}{d\theta} \left[sin\theta \frac{d(bP_1)}{d\theta}\right] = 0$ Since $b$ is a constant, it can be removed outside of the differential: $\frac{b\ sin\theta}{bP_1} \frac{d}{d\theta} \left[sin\theta \frac{dP_1}{d\theta}\right] = 0$ $\frac{sin\theta}{P_1} \frac{d}{d\theta} \left[sin\theta \frac{dP_1}{d\theta}\right] = 0$ Therefore, since $P_1$ is a solution, so is $bP_1$ for any non-zero constant $b$. If $F_1$ is a solution, then for $cF_1$: $\frac{1}{cF_1} \frac{d^2(cF_1)}{d\phi^2} = 0$ Since $c$ is a constant, it can be removed outside of the differential: $\frac{c}{cF_1} \frac{d^2F_1}{d\phi^2} = 0$ $\frac{1}{F_1} \frac{d^2F_1}{d\phi^2} = 0$ Therefore, since $F_1$ is a solution, so is $cF_1$ for any non-zero constant $c$. Task 2: If R is a constant, then $\frac{dR}{dr} = 0$ $\therefore \frac{1}{R} \frac{d}{dr}\left[0\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = l(l+1)$ $ -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = l(l+1)$ Since $r > 0$, and $l> 0$, this equation is not satisfied by a constant value of $R$. If P is a constant, then $\frac{dP}{d\theta} = m_l^2$ $\therefore \frac{sin\theta}{P} \frac{d}{d\theta} \left[0\right] + l(l+1)sin^2\theta= m_l^2$ $l(l+1)sin^2\theta = m_l^2$ The equation is not satisfied since there is still a dependence on $\theta$, which means that the wavefunction would only be permitted for certain values of $\theta$ Therefore a constant value of $P$ is not permissible. If F is a constant, then: $m_l^2 = 0$ $m_l = 0$ Task 3: a) Let $P = sin\theta$ $\therefore \frac{dP}{d\theta} = cos\theta$ $\rightarrow \frac{sin\theta}{sin\theta} \frac{d}{d\theta} \left[sin\theta cos\theta \right] + l(l+1)sin^2 \theta= m_l^2$ $cos(2\theta) + l(l+1)sin^2 \theta = m_l^2$ Expanding $cos (2\theta)$ as $cos^2\theta - sin^2\theta$, and simplifying gives: $cos^2\theta + (l^2 +l -1)sin^2\theta = m_l^2$ To eliminate the trigonometric terms, we wish to use the identity $cos^2\theta + sin^2\theta = 1$ $\therefore l^2 + l -2 = 0$ $(l+2)(l-1) = 1$ $\mathbf{ l =1}$ since $l > 0$ $\mathbf{m_l = \pm 1}$ b) Now let $P = cos\theta$ $\therefore \frac{dP}{d\theta} = -sin\theta$ $\frac{sin\theta}{cos\theta}\frac{d}{d\theta}\left[-sin^2\theta\right] + l(l+1)sin^2\theta = m_l^2$ Differentiating, simplifying and collecting terms yields: $(l^2 + l -2) sin^2\theta = m_l^2$ For this to be valid, $\mathbf{m_l^2 = 0}$, which means that either $sin^2 \theta$ or $(l^2 + l -2)$ must be equal to zero. Clearly, for the wavefunction to exist: $l^2+ 1 -2 = 0$ $\mathbf{l=1}$ since $l > 0$ c) Substituting in $P = tan\theta$ yields an equation still containing trigonometry. There is no way for this to be removed, and so the original trial solution is invalid. If you haven't tried any other trigonometric functions yet, why not try $cos^2\theta$, $sin(2\theta)$ and $sin^2\theta$...? Task 4: $\frac{1}{aR_1} \frac{d}{dr}\left[r^2 \frac{d(aR_1)}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = l(l+1)$ Given that $a_0 = \frac{\pi e^2 \mu}{h^2 \epsilon_0}$, $\alpha = \frac{e^2}{4\pi \epsilon_0}$, $E = -\frac{\mu e^4}{8n^2h^2\epsilon_0^2}$: $\frac{1}{aR_1} \frac{d}{dr}\left[r^2 \frac{d(aR_1)}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[-\frac{\mu e^4 r^2}{8n^2h^2\epsilon_0^2} +\frac{e^2r}{4\pi \epsilon_0}] = l(l+1)$ $\frac{1}{R}\frac{d}{dr} \left[r^2\frac{dR}{dr}\right] + (\frac{a_0r}{n})^2 -2a_0r = l(l+1)$ </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> This problem provides a glimpse into the way differential equations develop at university both in terms of mathematical structure and also in their applications. It is ideal for high achieving maths students coming to the end of their school careers who wish to draw their learning together or to prepare themselves for the university challenges ahead. <h3>Possible approach</h3> <div>This problem is most suitable for the highest attaining mathematics students or those with a very strong interest in pursuing mathematical physics.</div> <div>One thing might strike students: the problem involves a lot of reading and several parts to consider and solve. Encourage those working on it to read each part carefully and consider the mathematical ideas raised: the thinking about the issues can be of as much importance as doing the problem; reflecting on the solution obtained rather than simply getting the algebra correct is crucial! Furthermore, students who have become accustomed to completely solving problems might need to be reminded that partial solutions are acceptable and that at higher educational levels complete solutions and not the norm.</div> <div>It is often a good idea to encourage students to talk about their mathematical experiences. Perhaps this problem could be given to students who consider the problem with plenty of thinking space and then give a short presentation on the problem to the rest of the group at a later date. Alternatively, small groups of interested students could consider the problem collectively and try to make sense of the ideas, being proactive in helping themselves out of difficulty.</div> <h3>Key questions</h3> <div>The problem is well structured into different parts, so could be attempted by students independently. Some good questions to help students out of difficulty are:</div> <div>Are you clear as to what is constant and what is variable in each case?</div> <div>Have you understood the meaning of the coordinate system?</div> <div>Have you understood the meaning of the variables?</div> <div>Do you have a plan: do you know what you are trying to solve? What sort of maths might help you to solve it?</div> <h3>Possible extension</h3> <div>There is plenty of extension built into this problem. Interested students might also try more of the problems from the <a href="http://nrich.maths.org/6467&part=">Advanced Scientific Mathematics</a> section of stemNRICH.</div> <h3>Possible support</h3> <div>There are many more straightforward differential equation problems on the <a href="http://nrich.maths.org/6467&part=">Advanced Scientific Mathematics</a> section of stemNRICH.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This problem is tough and will require very careful algebraic manipulation. To have any chance of solving such equations requires you to look carefully at their structure and trial functional forms which have a chance of cancelling in the required way. You will need to know about differential equations to be able to solve this problem. </mdoxml> 2 5 0 0 0 0 1 The real hydrogen atom Dip your toe into the world of quantum mechanics by looking at the Schrodinger equation for hydrogen atoms Applications physics Pre-Calculus and Calculus 1st and 2nd order linear differential equations with constant coefficients Pre-Calculus and Calculus Differential equations Applications chemistry Applications physics Admin Individual Admin Individual Collections Fundamental Particles