Question

(a) A particle of mass $m$ moves in a spherical potential $V(r)$. Show that according to classical mechanics $$ \frac{\mathrm{d}}{\mathrm{d} t}\left(\mathbf{p} \times \mathbf{L}_{\mathrm{c}}\right)=m r^{2} \frac{\mathrm{d} V}{\mathrm{~d} r} \frac{\mathrm{d} \mathbf{e}_{r}}{\mathrm{~d} t} $$ where $\mathbf{L}_{\mathrm{c}}=\mathbf{r} \times \mathbf{p}$ is the classical angular-momentum vector and $\mathbf{e}_{r}$ is the unit vector in the radial direction. Hence show that when $V(r)=-K / r$, with $K$ a constant, the Runge-Lenz vector $\mathbf{M}_{\mathrm{c}} \equiv \mathbf{p} \times \mathbf{L}_{\mathrm{c}}-m K \mathbf{e}_{r}$ is a constant of motion. Deduce that $\mathbf{M}_{\mathrm{c}}$ lies in the orbital plane, and that for an elliptical orbit it points from the centre of attraction to the pericentre of the orbit, while it vanishes for a circular orbit. (b) Show that in quantum mechanics $(\mathbf{p} \times \mathbf{L})^{\dagger}-\mathbf{p} \times \mathbf{L}=-2 \mathbf{i p}$. Hence explain why in quantum mechanics we take the Runge-Lenz vector operator to be $$ \mathbf{M} \equiv \frac{1}{2} \hbar \mathbf{N}-m K \mathbf{e}_{r} \quad \text { where } \quad \mathbf{N} \equiv \mathbf{p} \times \mathbf{L}-\mathbf{L} \times \mathbf{p} $$ Explain why we can write down the commutation relation $\left[L_{i}, M_{j}\right]=$ $\mathrm{i} \sum_{k} \epsilon_{i j k} M_{k}$ (c) Explain why $\left[p^{2}, N\right]=0$ and why $[1 / r, \mathbf{p} \times \mathbf{L}]=[1 / r, \mathbf{p}] \times \mathbf{L}$ Hence show that $$ [1 / r, \mathbf{N}]=\mathrm{i}\left\{\frac{1}{r^{3}}\left(r^{2} \mathbf{p}-\mathbf{x} \mathbf{x} \cdot \mathbf{p}\right)-\left(\mathbf{p} r^{2}-\mathbf{p} \cdot \mathbf{x} \mathbf{x}\right) \frac{1}{r^{3}}\right\} $$ (d) Show that $$ \left[p^{2}, \mathbf{e}_{r}\right]=\mathrm{i} \hbar\left\{-\left(\mathbf{p} \frac{1}{r}+\frac{1}{r} \mathbf{p}\right)+\sum_{j}\left(p_{j} \frac{x_{j}}{r^{3}} \mathbf{x}+\mathbf{x} \frac{x_{j}}{r^{3}} p_{j}\right)\right\} $$ (e) Hence show that $[H, \mathbf{M}]=0 .$ What is the physical significance of this result? (f) Show that (i) $\left[M_{i}, L^{2}\right]=\mathrm{i} \sum_{j k} \epsilon_{i j k}\left(M_{k} L_{j}+L_{j} M_{k}\right)$, (ii) $\left[L_{i}, M^{2}\right]=$ 0, where $M^{2} \equiv M_{x}^{2}+M_{y}^{2}+M_{z}^{2} .$ What are the physical implications of these results? (g) Show that $$ \left[N_{i}, N_{j}\right]=-4 \mathrm{i} \sum_{u} \epsilon_{i j u} p^{2} L_{u} $$ and that $$ \left[N_{i},\left(\mathbf{e}_{r}\right)_{j}\right]-\left[N_{j},\left(\mathbf{e}_{r}\right)_{i}\right]=-\frac{4 \mathrm{i} \hbar}{r} \sum_{t} \epsilon_{i j t} L_{t} $$ and hence that $$ \left[M_{i}, M_{j}\right]=-2 \mathrm{i} \hbar^{2} m H \sum_{k} \epsilon_{i j k} L_{k} $$ What physical implication does this equation have?

Name: Problem 15, Chapter 8: The Physics of Quantum Mechanics
Uploaded: 2021-12-20T05:57:55-08:00
Duration: 16 min 18 s
Channel: Amit Srivastava

   (a) A particle of mass $m$ moves in a spherical potential $V(r)$. Show that according to classical mechanics
$$
\frac{\mathrm{d}}{\mathrm{d} t}\left(\mathbf{p} \times \mathbf{L}_{\mathrm{c}}\right)=m r^{2} \frac{\mathrm{d} V}{\mathrm{~d} r} \frac{\mathrm{d} \mathbf{e}_{r}}{\mathrm{~d} t}
$$
where $\mathbf{L}_{\mathrm{c}}=\mathbf{r} \times \mathbf{p}$ is the classical angular-momentum vector and $\mathbf{e}_{r}$ is the unit vector in the radial direction. Hence show that when $V(r)=-K / r$, with $K$ a constant, the Runge-Lenz vector $\mathbf{M}_{\mathrm{c}} \equiv \mathbf{p} \times \mathbf{L}_{\mathrm{c}}-m K \mathbf{e}_{r}$ is a constant of motion. Deduce that $\mathbf{M}_{\mathrm{c}}$ lies in the orbital plane, and that for an elliptical orbit it points from the centre of attraction to the pericentre of the orbit, while it vanishes for a circular orbit.
(b) Show that in quantum mechanics $(\mathbf{p} \times \mathbf{L})^{\dagger}-\mathbf{p} \times \mathbf{L}=-2 \mathbf{i p}$. Hence explain why in quantum mechanics we take the Runge-Lenz vector operator to be
$$
\mathbf{M} \equiv \frac{1}{2} \hbar \mathbf{N}-m K \mathbf{e}_{r} \quad \text { where } \quad \mathbf{N} \equiv \mathbf{p} \times \mathbf{L}-\mathbf{L} \times \mathbf{p}
$$
Explain why we can write down the commutation relation $\left[L_{i}, M_{j}\right]=$ $\mathrm{i} \sum_{k} \epsilon_{i j k} M_{k}$
(c) Explain why $\left[p^{2}, N\right]=0$ and why $[1 / r, \mathbf{p} \times \mathbf{L}]=[1 / r, \mathbf{p}] \times \mathbf{L}$ Hence show that
$$
[1 / r, \mathbf{N}]=\mathrm{i}\left\{\frac{1}{r^{3}}\left(r^{2} \mathbf{p}-\mathbf{x} \mathbf{x} \cdot \mathbf{p}\right)-\left(\mathbf{p} r^{2}-\mathbf{p} \cdot \mathbf{x} \mathbf{x}\right) \frac{1}{r^{3}}\right\}
$$
(d) Show that
$$
\left[p^{2}, \mathbf{e}_{r}\right]=\mathrm{i} \hbar\left\{-\left(\mathbf{p} \frac{1}{r}+\frac{1}{r} \mathbf{p}\right)+\sum_{j}\left(p_{j} \frac{x_{j}}{r^{3}} \mathbf{x}+\mathbf{x} \frac{x_{j}}{r^{3}} p_{j}\right)\right\}
$$
(e) Hence show that $[H, \mathbf{M}]=0 .$ What is the physical significance of this result?
(f) Show that (i) $\left[M_{i}, L^{2}\right]=\mathrm{i} \sum_{j k} \epsilon_{i j k}\left(M_{k} L_{j}+L_{j} M_{k}\right)$, (ii) $\left[L_{i}, M^{2}\right]=$ 0, where $M^{2} \equiv M_{x}^{2}+M_{y}^{2}+M_{z}^{2} .$ What are the physical implications of these results?
(g) Show that
$$
\left[N_{i}, N_{j}\right]=-4 \mathrm{i} \sum_{u} \epsilon_{i j u} p^{2} L_{u}
$$
and that
$$
\left[N_{i},\left(\mathbf{e}_{r}\right)_{j}\right]-\left[N_{j},\left(\mathbf{e}_{r}\right)_{i}\right]=-\frac{4 \mathrm{i} \hbar}{r} \sum_{t} \epsilon_{i j t} L_{t}
$$
and hence that
$$
\left[M_{i}, M_{j}\right]=-2 \mathrm{i} \hbar^{2} m H \sum_{k} \epsilon_{i j k} L_{k}
$$
What physical implication does this equation have?

The Physics of Quantum Mechanics

James Binney, David… 1st Edition

Chapter 8, Problem 15

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Solved by Expert Amit Srivastava

12/20/2021

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(a) A particle of mass $m$ moves in a spherical potential $V(r)$. Show that according to classical mechanics $$ \frac{\mathrm{d}}{\mathrm{d} t}\left(\mathbf{p} \times \mathbf{L}_{\mathrm{c}}\right)=m r^{2} \frac{\mathrm{d} V}{\mathrm{~d} r} \frac{\mathrm{d} \mathbf{e}_{r}}{\mathrm{~d} t} $$ where $\mathbf{L}_{\mathrm{c}}=\mathbf{r} \times \mathbf{p}$ is the classical angular-momentum vector and $\mathbf{e}_{r}$ is the unit vector in the radial direction. Hence show that when $V(r)=-K / r$, with $K$ a constant, the Runge-Lenz vector $\mathbf{M}_{\mathrm{c}} \equiv \mathbf{p} \times \mathbf{L}_{\mathrm{c}}-m K \mathbf{e}_{r}$ is a constant of motion. Deduce that $\mathbf{M}_{\mathrm{c}}$ lies in the orbital plane, and that for an elliptical orbit it points from the centre of attraction to the pericentre of the orbit, while it vanishes for a circular orbit. (b) Show that in quantum mechanics $(\mathbf{p} \times \mathbf{L})^{\dagger}-\mathbf{p} \times \mathbf{L}=-2 \mathbf{i p}$. Hence explain why in quantum mechanics we take the Runge-Lenz vector operator to be $$ \mathbf{M} \equiv \frac{1}{2} \hbar \mathbf{N}-m K \mathbf{e}_{r} \quad \text { where } \quad \mathbf{N} \equiv \mathbf{p} \times \mathbf{L}-\mathbf{L} \times \mathbf{p} $$ Explain why we can write down the commutation relation $\left[L_{i}, M_{j}\right]=$ $\mathrm{i} \sum_{k} \epsilon_{i j k} M_{k}$ (c) Explain why $\left[p^{2}, N\right]=0$ and why $[1 / r, \mathbf{p} \times \mathbf{L}]=[1 / r, \mathbf{p}] \times \mathbf{L}$ Hence show that $$ [1 / r, \mathbf{N}]=\mathrm{i}\left\{\frac{1}{r^{3}}\left(r^{2} \mathbf{p}-\mathbf{x} \mathbf{x} \cdot \mathbf{p}\right)-\left(\mathbf{p} r^{2}-\mathbf{p} \cdot \mathbf{x} \mathbf{x}\right) \frac{1}{r^{3}}\right\} $$ (d) Show that $$ \left[p^{2}, \mathbf{e}_{r}\right]=\mathrm{i} \hbar\left\{-\left(\mathbf{p} \frac{1}{r}+\frac{1}{r} \mathbf{p}\right)+\sum_{j}\left(p_{j} \frac{x_{j}}{r^{3}} \mathbf{x}+\mathbf{x} \frac{x_{j}}{r^{3}} p_{j}\right)\right\} $$ (e) Hence show that $[H, \mathbf{M}]=0 .$ What is the physical significance of this result? (f) Show that (i) $\left[M_{i}, L^{2}\right]=\mathrm{i} \sum_{j k} \epsilon_{i j k}\left(M_{k} L_{j}+L_{j} M_{k}\right)$, (ii) $\left[L_{i}, M^{2}\right]=$ 0, where $M^{2} \equiv M_{x}^{2}+M_{y}^{2}+M_{z}^{2} .$ What are the physical implications of these results? (g) Show that $$ \left[N_{i}, N_{j}\right]=-4 \mathrm{i} \sum_{u} \epsilon_{i j u} p^{2} L_{u} $$ and that $$ \left[N_{i},\left(\mathbf{e}_{r}\right)_{j}\right]-\left[N_{j},\left(\mathbf{e}_{r}\right)_{i}\right]=-\frac{4 \mathrm{i} \hbar}{r} \sum_{t} \epsilon_{i j t} L_{t} $$ and hence that $$ \left[M_{i}, M_{j}\right]=-2 \mathrm{i} \hbar^{2} m H \sum_{k} \epsilon_{i j k} L_{k} $$ What physical implication does this equation have?

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Key Concepts

Central Potentials

A central potential is one that depends solely on the distance from a central point, typically written as V(r). This spherical symmetry implies that the force always points radially and that no direction in space is preferred. In both classical and quantum mechanics, the spherical symmetry of the potential leads to the conservation of angular momentum, which plays a fundamental role in determining the motion and allowed quantum states of a particle subject to such a force.

Angular Momentum Conservation

Angular momentum is a vector quantity conserved in systems with rotational symmetry, such as those involving a central potential. In classical mechanics, the angular momentum L = r × p remains constant in a central force field. Its conservation restricts the motion of the particle to a plane, and in quantum mechanics, the angular momentum operators satisfy well?defined commutation relations that underlie the rotational symmetry of the system and help determine energy level structures and selection rules.

Runge-Lenz Vector

The Runge-Lenz vector is an additional conserved quantity found in systems with an inverse-square central force, such as the Kepler problem. It is not associated with a continuous symmetry in the usual sense but reflects a hidden symmetry of the inverse-square law. In the classical context, it determines the orientation of an elliptical orbit (pointing from the centre of force to the pericentre) and in the quantum mechanical context, it is related to the degeneracy of bound states and helps in formulating an underlying dynamical symmetry group.

Quantum Mechanical Operators and Hermiticity

In quantum mechanics, physical observables are represented by operators that must be Hermitian to ensure real eigenvalues. When formulating the quantum analog of quantities like the Runge-Lenz vector, issues of operator ordering and the requirement of Hermiticity become crucial. Adjustments, such as symmetrizing the product of non-commuting operators, are necessary to construct meaningful operators that respect the physical properties of the corresponding classical variables.

Commutation Relations and Lie Algebras

Commutation relations are fundamental in quantum mechanics as they encode the underlying algebraic structure of the observables. For example, the angular momentum operators obey specific commutation rules that form the Lie algebra of the rotation group. Similarly, the commutation relations involving the Runge-Lenz vector reveal a hidden so(4) symmetry in the Kepler problem. This algebraic structure is crucial for understanding the degeneracy of energy levels and the symmetry properties of quantum systems.

Hidden Symmetry and Conserved Quantities

Hidden symmetries refer to conservation laws that are not immediately apparent from the spatial symmetries of the system but emerge from the dynamics, such as the conservation of the Runge-Lenz vector in the Kepler problem. These hidden symmetries lead to unexpected degeneracies in the energy spectrum and provide deeper insight into the dynamical behavior of systems governed by inverse-square laws, both in classical trajectories and quantum mechanical bound states.

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Transcript

00:02 An orbiting charged particle radiates the electromagnetic energy and for an electron in atomic dimensions it would lead to collapse in considerably less than the wing of an eye.

00:31 For first by equating the centripetal and the column forces show that for a classical charge of mass m held in the circular orbit by its attraction to a fixed charge of plus q or plus e, the following relationship holds.

01:06 That is omega is equal to e r to the power minus 3 by 2 upon 4 pi epsilon not n and for b part the electromagnetism tells us that a charge whose acceleration is a charge has acceleration a radiates the power p equals to e square a square by 6 epsilon not c cube show that this can also be expressed in terms of the orbital radius that is e to the power 6 by 96 pi square epsilon not cube by m cq while to the power 4 then calculate the energy lost per orbit energy lost per orbit in terms of r by multiplying this power by the time period which is equal to 2 pi by omega and using the formula from the part a to eliminate the omega and for c part in such a classical orbit the total mechanical energy total mechanical energy is half the potential energy or we can say that e orbit is equal to minus of e square by 8 pi epsilon not r calculate the change in energy change in energy per change in r that is d .e by dr.

03:38 From this rate of change and the energy loss per orbit from part b determine the change in r per orbit and evaluate it for a typical orbit of 10 to the power minus 10 meter would the electrons radius change much in a single orbit and for d argue that dividing d by dr by p and multiplying it by dr gives the time required for r to change by dr.

04:29 Then sum this time for all the radii for all radii from from r orbital to final radius of 0 and evaluate the result for the given orbital radius 10 to minus 10 meter.

04:58 For first part, the column 4 is given by minus of e square by 4x0, r squared.

05:08 Here e is the electron charge, epsilon not is the permittivity of free space, r is the electron distance from the nucleus set this column force equals to the centricidal force which is required to maintain the circular motion and we are using here the relation of the velocity is angular velocity multiplied by the radial distance to obtain the electron mass so this will be minus e square by 4 pi a epsilon not r square is equal to the centripetal force is m -a which can be written as minus of m v square by r where we can also be written as omega into r whole square this is equal to minus m omega square r now simplifying the equation for omega square is equal to e square by four pi a xcelon not m r q so the omega or angular velocity will be equals to e r to the power minus 3 by 2 divided by under root of 4 pi epsilon not and this is the angular frequency of the particle.

06:40 Now for the second part the acceleration a is the centripetal acceleration which is equals to centripetal acceleration contritoral acceleration is equal to omega square r as in party but we have to substitute this acceleration in the given equation of power which is e square a square by 6 absalom not c cube this is equal to e square omega square r whole square divided by 6 absalom not c cube now substituting the angle of frequency derived previously, so this will be equals to e square, e square by 4 pi epsilon 0 mr q into r whole square.

07:47 This is divided by 6 epsilon not c -qu.

07:50 On simplification, the power is equal to 8 the power 6 divided by 19.

07:59 5x2, epsilon 0, x2, um square, m square c cube, r to the power 4, where the changing energy of the orbit is equals to minus of power multiplied by the time period, substituting the value, e to the power 6, minus of e to the power 6 divided by 96 pi square, xb0 .m squared c q r to the power 4 multiplied by 2 pi multiplied by under root 4 pi epsilon not m by e r to the power minus 3 by 2.

08:56 On simplification the energy change will be equal to minus of e to the power 5 r to the power minus o 5 5 5 by 2.

09:08 Divide by 24 root 5 epsilon 0 5 by 2 m to the power 3 by 2 ccc and 4 the third part starting from the energy of the electron the energy of the orbit is given by energy of orbit e orbit is equal to minus e square by 8 5 xxon 0 r we have to differentiate this energy with respect to r this is equal to e square by 8 pi a xxel not r square then multiplied it by del r so the del r is equal to d e by d r by d r of the orbit inverse multiplied by the change in energy of the orbit.

10:31 So this is equal to 8x epsilon not r square by e square into the energy change of the orbit.

10:40 Using the energy change in four single orbit to find the expression for change in radius is equal to 8 phi epsilon not r square by e square multiplied by minus of e to the power 5 divided by 24, 5 to the power 3 by 2, epsilon 0, west to the power 5 by 2, n to the power 3 by 2, c cube r to the power 5 by 2...

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