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

Question

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Answer 1

Question

Chapter 4, Problem 4.65P

To determine

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

Expert Solution & Answer

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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Students have asked these similar questions

For number 11 please sketch the harmonic on graphing paper.

# E 94 20 13. Time a) What is the frequency of the above wave? b) What is the period? c) Highlight the second cycle d) Sketch the sine wave of the second harmonic of this wave % 7 & 5 6 7 8 * ∞ Y U 9 0 0 P 150

Show work using graphing paper

Answer 2

Question

Chapter 4, Problem 4.65P

To determine

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

Expert Solution & Answer

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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Answer 3

Question

Chapter 4, Problem 4.65P

To determine

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

Expert Solution & Answer

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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Answer 4

Question

Chapter 4, Problem 4.65P

To determine

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

Expert Solution & Answer

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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Answer 5

Chapter 4, Problem 4.65P

To determine

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

Expert Solution & Answer

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|↓↓↑〉)

Answer 6

Chapter 4, Problem 4.65P

To determine

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

Expert Solution & Answer

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|↓↓↑〉)

Answer 7

Chapter 4, Problem 4.65P

To determine

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

Answer 8

Expert Solution & Answer

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|↓↓↑〉)

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

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|↓↓↑〉)

Answer 12

Ch. 4.1 - Prob. 4.1P Ch. 4.1 - Prob. 4.3P Ch. 4.1 - Prob. 4.4P Ch. 4.1 - Prob. 4.5P Ch. 4.1 - Prob. 4.6P Ch. 4.1 - Prob. 4.7P Ch. 4.1 - Prob. 4.8P Ch. 4.1 - Prob. 4.9P Ch. 4.1 - Prob. 4.10P Ch. 4.1 - Prob. 4.11P

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

Concept explainers

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

Answer to Problem 4.65P

Explanation of Solution

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