Q57P The fundamental commutation rela... [FREE SOLUTION]

91影视

Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 4: Q57P (page 197)

The fundamental commutation relations for angular momentum (Equation 4.99) allow for half-integer (as well as integer) eigenvalues. But for orbital angular momentum only the integer values occur. There must be some extra constraint in the specific form $L = r 脳 p$ that excludes half-integer values. Let be some convenient constant with the dimensions of length (the Bohr radius, say, if we're talking about hydrogen), and define the operators
$q_{1} 鈮� \frac{1}{\sqrt{2}} [x + (a^{2} / 魔) p_{y}]; p_{1} 鈮� \frac{1}{\sqrt{2}} [p_{x} - (魔 / a^{2}) y]; q_{2} 鈮� \frac{1}{\sqrt{2}} [x - (a^{2} / 魔) p_{y}]; p_{2} 鈮� \frac{1}{\sqrt{2}} [p_{x} - (魔 / a^{2}) y];$
(a) Verify that $[q_{1,} q_{2}] = [p_{1}, p_{2}] = 0; [q_{1,} p_{1}] = [p_{2}, q_{2}] = i 魔$ . Thus the q's and the p's satisfy the canonical commutation relations for position and momentum, and those of index 1are compatible with those of index 2 .
(b) Show that $[q_{1,} q_{2}] = [p_{1}, p_{2}] L_{z} = \frac{魔}{2 a^{2}} (q_{1}^{2} - q_{2}^{2}) + \frac{a^{2}}{2 魔} (p_{1}^{2} - p_{2}^{2})$
(c) Check that , where each is the Hamiltonian for a harmonic oscillator with mass and frequency .
(d) We know that the eigenvalues of the harmonic oscillator Hamiltonian are , where (In the algebraic theory of Section this follows from the form of the Hamiltonian and the canonical commutation relations). Use this to conclude that the eigenvalues of must be integers.

Short Answer

Expert verified

a) The equations are verified.

b) It is proved that.

c) The equationis checked and verified.

d) It is concluded that the eigenvalues of are integers.

Step by step solution

Definition of angular momentum

The rotating equivalent of linear momentum is angular momentum. Because it is a conserved quantity鈥攖he total angular momentum of a closed system remains constant鈥攊t is a significant quantity in physics.

Both the direction and the amplitude of angular momentum are preserved.

Verify that [q1,q2]=[p1,p2]=0;[q1,q2]=[q2,p2]=iħ

(a)

Consider that,

$q_{1} 鈮� \frac{1}{\sqrt{2}} [x + (a^{2} / 魔) p_{y}]; q_{1} 鈮� \frac{1}{\sqrt{2}} [p x - {(魔 / a^{2})}_{y}]; p_{2} 鈮� \frac{1}{\sqrt{2}} [x - (a^{2} / 魔) p_{y}]; q_{2} 鈮� \frac{1}{\sqrt{2}} [p x + {(魔 / a^{2})}_{y}];$

Now, the commutation relation between $q_{1}$ and $q_{2_{}}$ is,

localid="1658311027712" $[q_{1}, q_{2}] = [\frac{1}{\sqrt{2}} (x + (\frac{a^{2}}{魔}) p_{y}), \frac{1}{\sqrt{2}} (x - (\frac{a^{2}}{魔}) p_{y})] = \frac{1}{2} [[x, x] + x, (- \frac{a^{2}}{魔}) p_{y}] + [(\frac{a^{2}}{魔}) p_{y}, (\frac{a^{2}}{魔}) p_{y}] = 0$

Since, $[x, x] = [p_{y}, p_{y}] = 0$

Also, the commutation relation between $p_{1}$ and $p_{2}$ is,

localid="1658311843930" $[p_{1}, p_{2}] = [\frac{1}{\sqrt{2}} (p_{x} - (\frac{魔}{a^{2}}) y), \frac{1}{\sqrt{2}} (p_{x} + (\frac{魔}{魔 a^{2}}) y)] = \frac{1}{2} [p_{x}, p_{x}] + \frac{1}{2} [p_{x}, \frac{魔}{a^{2}} y] - \frac{1}{2} [\frac{魔}{a^{2}} y, p_{x}] - \frac{1}{2} [\frac{魔}{a^{2}} y, \frac{a^{2}}{魔} y] A s, [p_{x}, p_{x}] = [y, y] = [p_{x}, y] = 0$

The commutation relation between $q_{1}$ and $p_{1}$ is,

role="math" localid="1658313061062" $[q_{1}, p_{1}] = [\frac{1}{\sqrt{2}} (x + \frac{a^{2}}{魔}), \frac{1}{\sqrt{2}} (p_{x} - (\frac{魔}{a^{2}}) y)] = \frac{1}{2} [x, p_{x}] + \frac{1}{2} [\frac{a^{2}}{魔} p_{y}, p_{x}] - \frac{1}{2} [x, \frac{魔}{a^{2}} y] - \frac{1}{2} [\frac{a^{2}}{魔} p_{y}, \frac{魔}{a^{2}}] = \frac{1}{2} [x, p_{x}] - \frac{1}{2} [p_{y}, y] = \frac{1}{2} [i 魔 - (- i 魔)] = \frac{2 i 魔}{2} = i 魔$

And, commutation relation between $q_{2}$ and $p_{2}$ is,

role="math" localid="1658313694232" $[q_{2}, p_{2}] = [\frac{1}{\sqrt{2}} (x - \frac{a^{2}}{魔} p_{y}), \frac{1}{\sqrt{2}} (p_{x} + \frac{魔}{a^{2}} y)] = \frac{1}{2} [x, p_{x}] + \frac{1}{2} [x, \frac{魔}{a^{2}} y] - \frac{1}{2} [\frac{a^{2}}{魔} p_{y}, p_{x}] - \frac{1}{2} [\frac{a^{2}}{魔} p_{y}, \frac{魔}{a^{2}} y] = \frac{1}{2} [x, p_{x}] - \frac{1}{2} [p_{y}, y] = \frac{2 i 魔}{2} = i 魔 A s, [x, y] = [p_{y}, p_{x}] = 0 B u t [x, p_{x}] = i 魔 a n d [p_{y}, y] = - i 魔 T h e r e f o r e, t h e a b o v e f i n d i n g s u g g e s t s t h a t q' s a n d p' s f o l l o w t h e p o s i t i o n - m o m e n t u m q u a n t u m c o m m u t a t i o n r e l a t i o n s, [q_{i}, p_{j}] = i 魔未_{i j} .$

Show that Lz=ħ2a2(q12-q22)+a22ħ(p12-p22)

(b)

Evaluate $q_{1}^{2} - q_{2}^{2}$ and $p_{1}^{2} - p_{2}^{2}$ to obtain their relation with the z-component of the angular momentum:

$q_{1}^{2} - q_{2}^{2} = \frac{1}{2} \{x^{2} + (\frac{a^{2}}{魔}) p_{y}^{2} + (\frac{a^{2}}{魔}) \{x, p_{y}\}\} - \frac{1}{2} \{x^{2} + (\frac{a^{2}}{魔}) p_{y}^{2} + (\frac{a^{2}}{魔}) \{x, p_{y}\}\} = (\frac{a^{2}}{魔}) \{x, p_{y}\} = 2 (\frac{a^{2}}{魔}) x p_{y}$

Also,

$p_{1}^{2} - p_{2}^{2} = \frac{1}{2} [p_{x}^{2} + {(\frac{魔}{a^{2}})}^{2} y^{2} - \frac{魔}{a^{2}} (p_{x} y + y p_{x}) - p_{x}^{2} - {(\frac{魔}{a^{2}})}^{2} y^{2} - \frac{魔}{a^{2}} (p_{x} y + y p_{x})] = - \frac{2 魔}{a^{2}} y p_{x}$

Now,

$\frac{魔}{2 a^{2}} (q_{1}^{2} - q_{2}^{2}) + \frac{a^{2}}{2 魔} (p_{1}^{2} - p_{2}^{2}) = \frac{魔}{2 a^{2}} (\frac{2 a^{2}}{魔} x p_{y}) + \frac{a^{2}}{2 魔} (\frac{2 魔}{a^{2}} x p_{x}) = x p_{y} - y p_{x} = L_{z}$

Thus, the relation with the z component of angular momentum is established.

Check that Lz=H1-H2

(c)

The Hamiltonian for the harmonic oscillator is,

$H = \frac{p^{2}}{2 m} + \frac{1}{2} m 蝇^{2} x^{2} For m = \frac{魔}{a^{2}} and 蝇 = 1 H = \frac{1}{2 m} P^{2} + \frac{1}{2} {m蝇}^{2} x^{2} = \frac{a^{2}}{2 魔} p^{2} + \frac{魔}{2 a^{2}} x^{2}$

And

$H_{1} = \frac{a^{2}}{2 魔} p_{1}^{2} + \frac{魔}{2 a^{2}} q_{1}^{2} H_{2} = \frac{a^{2}}{2 魔} p_{2}^{2} + \frac{魔}{2 a^{2}} q_{2}^{2} H_{1} - H_{2} = \frac{a^{2}}{2 魔} (p_{1}^{2} - p_{2}^{2}) + \frac{魔}{2 a^{2}} (q_{1}^{2} - q_{2}^{2}) = L_{z}$

Thus, the relationship is established.

The eigenvalues of Lz must be integers

(d)

The values of eigenvalues of $L_{z}$ are integers. So,

$H 蠄鈱� = (n - \frac{1}{2}) 魔蝇$

Then,

$(H_{1} - H_{2}) 蠄鈱� = L_{z} 蠄鉄� = (n_{1} - \frac{1}{2} - n_{2} - \frac{1}{2}) 魔蝇蠄鈱�$

For $蝇 = 1$ :

$L_{2} 蠄鈱� = (n_{1}, n_{2}) 魔蠄鉄�$

Here $n_{1}, n_{2} = 0, 1, 2, 3, . . .$

$L_{z} 蠄鈱� = m 魔蠄鉄�,$ is an integer since .

Hence, the eigenvalues of $L_{z}$ must be integers.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Short Answer

Step by step solution

Definition of angular momentum

Verify that [q1,q2]=[p1,p2]=0;[q1,q2]=[q2,p2]=iħ

Show that Lz=ħ2a2(q12-q22)+a22ħ(p12-p22)

Check that Lz=H1-H2

The eigenvalues of Lz must be integers

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Electromagnetism

Modern Physics

Magnetism

Radiation

Famous Physicists

Study anywhere. Anytime. Across all devices.