Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
3rd Edition
ISBN: 9781107189638
Author: Griffiths, David J., Schroeter, Darrell F.
Publisher: Cambridge University Press
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Connection formulae for WKB approximation
In quantum mechanics, Schrödinger’s equation for certain systems with varying parameters cannot be solved directly. In such a case, an approximation method is used to obtain the approximate solution. WKB method is a special technique used to obtain the a…
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Chapter 4, Problem 4.65P
To determine

Construct the quadruplet and the two doublets using the notation of Equation 4.175 and 4.176.

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Answer to Problem 4.65P

Constructed the quadruplet and the two doublets using the notation of Equation 4.175 and 4.176:

The quadruplet is,

|3232〉=|↑↑↑〉|3212〉=13(|↓↑↑〉+|↑↓↑〉+|↑↑↓〉)|32−12〉=13(|↓↓↑〉+|↓↑↓〉+|↑↓↓〉)|32−32〉=|↓↓↓〉

The doublet 1 is

|1212〉1=12(|↑↓↑〉−|↓↑↑〉)|12−12〉1=12|↑↓↓〉−|↓↑↓〉

The doublet 2 is

|1212〉2=16(−2|↑↑↓〉+|↑↓↑〉+|↓↑↑〉)|12−12〉2=16(−|↓↑↓〉−|↑↓↓〉+2|↓↓↑〉)

Explanation of Solution

To construct the quadruplet:

Let |3232〉=|↑↑↑〉

Write the expression for lowering operator for one particle, Equation 4.146

    S−|↑〉=ℏ|↓〉

And,

    S−|↓〉=0

For all three states,

S=S1+S2+S3

Therefore, the other states of the lowering operator is

S−=S1−+S2−+S3−

From above equations,

S−|3232〉=S−|↑↑↑〉=ℏ(|↓↑↑〉+|↑↓↑〉+|↑↑↓〉)=B3/23/2|3212〉=3ℏ|3212〉

|3212〉=13(|↓↑↑〉+|↑↓↑〉+|↑↑↓〉)

Solving to find |32−12〉.

S−|3212〉=ℏ3(|↓↓↑〉+|↓↑↓〉+|↓↓↑〉+|↑↓↓〉+|↓↑↓〉+|↑↓↓〉)=2ℏ3(|↓↓↑〉+|↓↑↓〉+|↑↓↓〉)=B3/21/2|32−12〉|32−12〉=13(|↓↓↑〉+|↓↑↓〉+|↑↓↓〉)

Solving to find |32−32〉.

S−|32−12〉=ℏ3(|↓↓↓〉+|↓↓↓〉+|↓↓↓〉)=3ℏ|↓↓↓〉=B3/2−1/2|32−32〉

Solving further,

S−|32−12〉=3ℏ|32−32〉|32−32〉=|↓↓↓〉

Thus, the quadruplet is

|3232〉=|↑↑↑〉|3212〉=13(|↓↑↑〉+|↑↓↑〉+|↑↑↓〉)|32−12〉=13(|↓↓↑〉+|↓↑↓〉+|↑↓↓〉)|32−32〉=|↓↓↓〉

To construct doublet 1:

Let |1212〉1=12(|↑↓↑〉−|↓↑↑〉).

S−|1212〉1=ℏ2(|↓↓↑〉+|↑↓↓〉−|↓↓↑〉−|↓↑↓〉)=ℏ2(|↑↓↓〉−|↓↑↓〉)=B1/21/2|12−12〉1=ℏ|12−12〉1

Hence,

|12−12〉1=12|↑↓↓〉−|↓↑↓〉

Thus, the doublet 1 is

|1212〉1=12(|↑↓↑〉−|↓↑↑〉)|12−12〉1=12|↑↓↓〉−|↓↑↓〉

To construct doublet 2:

Let |1212〉2=a|↑↑↓〉+b|↑↓↑〉+c|↓↑↑〉 (orthogonal to |1212〉1 and |3212〉).

1〈1212|1212〉2=12(〈↑↓↑|−〈↓↑↑|)(a|↑↑↓〉+b|↑↓↑〉+c|↓↑↑〉)=12(b−c)=0

Since,

b−c=0b=c

Similarly to find a&b,

〈3212|1212〉2=13(〈↓↑↑|+〈↑↓↑|+〈↑↑↓|)(a|↑↑↓〉+b|↑↓↑〉+c|↓↑↑〉)=13(a+b+c)=0

Since,

a+b+c=0a+2b=0a=−2b

For normalization,

|a|2+|b|2+|c|2=1

Substituting the relation of a,b,c in the above equation,

4|b|2+|b|2+|b|2=16|b|2=1|b|2=16b=16

The, c=16 and 1=−26.

Substituting the value of a,b&c in |1212〉2=a|↑↑↓〉+b|↑↓↑〉+c|↓↑↑〉

|1212〉2=−26|↑↑↓〉+16|↑↓↑〉+16|↓↑↑〉=16(−2|↑↑↓〉+|↑↓↑〉+|↓↑↑〉)

To solve for |12−12〉2

S−|1212〉2=ℏ6(−2|↓↑↓〉−2|↑↓↓〉+|↓↓↑〉+|↑↓↓〉+|↓↓↑〉+|↓↑↓〉)=ℏ6(−|↓↑↓〉−|↑↓↓〉+2|↓↓↑〉)=B1/21/2|1212〉2=ℏ|12−12〉2

Therefore,

|12−12〉2=16(−|↓↑↓〉−|↑↓↓〉+2|↓↓↑〉)

Thus, the doublet 2 is

|1212〉2=16(−2|↑↑↓〉+|↑↓↑〉+|↓↑↑〉)|12−12〉2=16(−|↓↑↓〉−|↑↓↓〉+2|↓↓↑〉)

Conclusion:

Constructed the quadruplet and the two doublets using the notation of Equation 4.175 and 4.176:

The quadruplet is

|3232〉=|↑↑↑〉|3212〉=13(|↓↑↑〉+|↑↓↑〉+|↑↑↓〉)|32−12〉=13(|↓↓↑〉+|↓↑↓〉+|↑↓↓〉)|32−32〉=|↓↓↓〉

The doublet 1 is

|1212〉1=12(|↑↓↑〉−|↓↑↑〉)|12−12〉1=12|↑↓↓〉−|↓↑↓〉

The doublet 2 is

|1212〉2=16(−2|↑↑↓〉+|↑↓↑〉+|↓↑↑〉)|12−12〉2=16(−|↓↑↓〉−|↑↓↓〉+2|↓↓↑〉)

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Chapter 4 Solutions

Introduction To Quantum Mechanics

Ch. 4.1 - Prob. 4.1PCh. 4.1 - Prob. 4.3PCh. 4.1 - Prob. 4.4PCh. 4.1 - Prob. 4.5PCh. 4.1 - Prob. 4.6PCh. 4.1 - Prob. 4.7PCh. 4.1 - Prob. 4.8PCh. 4.1 - Prob. 4.9PCh. 4.1 - Prob. 4.10PCh. 4.1 - Prob. 4.11P
Ch. 4.2 - Prob. 4.12PCh. 4.2 - Prob. 4.13PCh. 4.2 - Prob. 4.14PCh. 4.2 - Prob. 4.15PCh. 4.2 - Prob. 4.16PCh. 4.2 - Prob. 4.17PCh. 4.2 - Prob. 4.18PCh. 4.2 - Prob. 4.19PCh. 4.2 - Prob. 4.20PCh. 4.3 - Prob. 4.21PCh. 4.3 - Prob. 4.22PCh. 4.3 - Prob. 4.23PCh. 4.3 - Prob. 4.24PCh. 4.3 - Prob. 4.25PCh. 4.3 - Prob. 4.26PCh. 4.3 - Prob. 4.27PCh. 4.4 - Prob. 4.28PCh. 4.4 - Prob. 4.29PCh. 4.4 - Prob. 4.30PCh. 4.4 - Prob. 4.31PCh. 4.4 - Prob. 4.32PCh. 4.4 - Prob. 4.33PCh. 4.4 - Prob. 4.34PCh. 4.4 - Prob. 4.35PCh. 4.4 - Prob. 4.36PCh. 4.4 - Prob. 4.37PCh. 4.4 - Prob. 4.38PCh. 4.4 - Prob. 4.39PCh. 4.4 - Prob. 4.40PCh. 4.4 - Prob. 4.41PCh. 4.5 - Prob. 4.42PCh. 4.5 - Prob. 4.43PCh. 4.5 - Prob. 4.44PCh. 4.5 - Prob. 4.45PCh. 4 - Prob. 4.46PCh. 4 - Prob. 4.47PCh. 4 - Prob. 4.48PCh. 4 - Prob. 4.49PCh. 4 - Prob. 4.50PCh. 4 - Prob. 4.51PCh. 4 - Prob. 4.52PCh. 4 - Prob. 4.53PCh. 4 - Prob. 4.54PCh. 4 - Prob. 4.55PCh. 4 - Prob. 4.56PCh. 4 - Prob. 4.57PCh. 4 - Prob. 4.58PCh. 4 - Prob. 4.59PCh. 4 - Prob. 4.61PCh. 4 - Prob. 4.62PCh. 4 - Prob. 4.63PCh. 4 - Prob. 4.64PCh. 4 - Prob. 4.65PCh. 4 - Prob. 4.66PCh. 4 - Prob. 4.70PCh. 4 - Prob. 4.72PCh. 4 - Prob. 4.73PCh. 4 - Prob. 4.75PCh. 4 - Prob. 4.76P
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