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

Consider a system of two particles, one with spin 1 and the other with spin 1/2. The particles interact with one another via the Hamiltonian H = Eo [S(1). S(2)] + E₁ [ S:(1) + S. (2)] ħ " where Eo and E₁ are constants with units of energy, S is the spin operator and the labels (1) and (2) indicate the particle on which it operates. (a) Find the energy spectrum of this system, i.e. the eigenvalues of the Hamiltonian H and identify the ground state and first excited state. For numerical purposes assume Eo = 0.2eV and E₁ = 0.1eV. (b) If the two particle system is known to be in a configuration such that the total spin is 1/2, what values a mea- surement of S% of the spin 1 particle would produce and what are the probabilities associated with each one of those values?

The Hamiltonian of a two-state system is given by Hˆ = H11 |1ih1| + H22 |2ih2| + H12 |1ih2| + H21 |2ih1| where the states |1i, |2i form an orthonormal basis. Since Hˆ must be Hermitian, H11 and H22 are real, while the complex (off-diagonal) elements satisfy H∗ 21 = H12. Find the eigenvalues of Hˆ . Find also the eigenvectors of Hˆ (as linear combinations of the states |1i and |2i).

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

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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