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

Question

Want to see more full solutions like this?

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|↓↓↑〉)

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Consider a composite state of spin j1 = s = 1/2 and angular momentum j2 = l = 2 of an electron. Find all the eigenstates of |j1, j2; j,m〉 as the linear combination of product states of spin and angular momentum. Give the values of Clebsch-Gordon coefficients you get from here.

Consider a particle of spin s = 3/2. (a) Find the matrices representing the operators S^ x , S^ y ,S^ z , ^ Sx 2 and ^ S y 2 within the basis of ^ S 2 and S^ z (b) Find the energy levels of this particle when its Hamiltonian is given by ^H= ϵ 0 h 2 ( Sx 2−S y 2 )− ϵ 0 h ( S^ Z ) where ϵ 0 is a constant having the dimensions of energy. Are these levels degenerate? (c) If the system was initially in an eigenstate Ψ0=( 1 0 0 0) , find the state of the system at time

Evaluate the spin matrices Sy and Szfor a particle with spin s = 1/2

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|↓↓↑〉)

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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|↓↓↑〉)

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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|↓↓↑〉)

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Want to see more full solutions like this?

Chapter 4 Solutions