Problem 3: Rotating a spin state (3 parts , 15 point)
PART A (6 points possible)
Consider the |+⟩ state of a spin one-half particle (as usual |±⟩=|z;±⟩) and apply to it the rotation operator R that rotates states by an angle θ around the y axis. Write out the resulting state |Ψ1⟩ in terms of the basis states |±⟩. Do not discard any phase.
Enter 'phi' for ϕ, 'theta' for θ, 'ketp' for |+⟩ and 'ketm' for |−⟩ as needed.
|Ψ1⟩=
PART B (6 points possible)
To the resulting state |Ψ1⟩ apply the rotation operator R that rotates by an angle ϕ around the z axis. Write out the resulting state |Ψ2⟩ in terms of the basis states |±⟩. Do not discard any phase.
Enter 'phi' for ϕ, 'theta' for θ, 'ketp' for |+⟩ and 'ketm' for |−⟩ as needed.
|Ψ2⟩=
PART C (3 points possible)
How is |Ψ2⟩ related to the state |n;+⟩ where n is the unit vector pointing in the (θ,ϕ) direction?
Enter 'phi' for ϕ, 'theta' for θ, 'ketp' for |+⟩ and 'ketm' for |−⟩ as needed.
|Ψ2⟩=
- unanswered
|n;+⟩
To solve these problems, we need to understand the concept of spin states and rotation operators.
PART A:
In this part, we need to apply the rotation operator R to the |+⟩ state and write out the resulting state |Ψ1⟩ in terms of the basis states |±⟩.
The rotation operator around the y-axis is given by:
R = e^(-iθS_y)
where S_y is the spin operator along the y-axis.
To find |Ψ1⟩, we need to apply the rotation operator R to the |+⟩ state:
|Ψ1⟩ = R|+⟩
To calculate this, we can use the formula for the rotation operator:
R|+⟩ = e^(-iθS_y)|+⟩
Now, we need to express the spin operators in terms of the basis states |±⟩. The spin operator along the y-axis can be written as:
S_y = (1/2)(|+⟩⟨-| - |-⟩⟨+|)
Substituting this into the expression for R, we get:
|Ψ1⟩ = e^(-iθS_y)|+⟩ = e^(-iθ(1/2)(|+⟩⟨-| - |-⟩⟨+|))|+⟩
Expand this expression and use the fact that |+⟩⟨-| |+⟩ = 0 and |-⟩⟨+| |+⟩ = |+⟩:
|Ψ1⟩ = e^(-iθ/2)(|+⟩ - |-⟩)
So, the resulting state |Ψ1⟩ in terms of the basis states |±⟩ is:
|Ψ1⟩ = e^(-iθ/2)|+⟩ - e^(-iθ/2)|-⟩
PART B:
In this part, we need to apply the rotation operator R (rotating by angle ϕ around the z-axis) to the resulting state |Ψ1⟩ from Part A and write out the resulting state |Ψ2⟩ in terms of the basis states |±⟩.
The rotation operator around the z-axis is given by:
R = e^(-iϕS_z)
where S_z is the spin operator along the z-axis.
To find |Ψ2⟩, we need to apply the rotation operator R to the |Ψ1⟩ state:
|Ψ2⟩ = R|Ψ1⟩
Using the formula for the rotation operator, we have:
R|Ψ1⟩ = e^(-iϕS_z)|Ψ1⟩
Now, we need to express the spin operator S_z in terms of the basis states |±⟩. The spin operator along the z-axis can be written as:
S_z = (1/2)(|+⟩⟨+| - |-⟩⟨-|)
Substituting this into the expression for R, we get:
|Ψ2⟩ = e^(-iϕS_z)|Ψ1⟩ = e^(-iϕ/2)(|+⟩⟨+| - |-⟩⟨-|)(e^(-iθ/2)|+⟩ - e^(-iθ/2)|-⟩)
Expand this expression and use the fact that ⟨+|+⟩ = ⟨-|-⟩ = 1, ⟨+|-⟩ = ⟨-|+⟩ = 0, we simplify the terms to:
|Ψ2⟩ = (e^(-iϕ/2)e^(-iθ/2))|+⟩ + (e^(-iϕ/2)e^(-iθ/2))|-⟩
Simplifying further, we get:
|Ψ2⟩ = e^(-i(ϕ/2 + θ/2))|+⟩ + e^(-i(ϕ/2 + θ/2))|-⟩
So, the resulting state |Ψ2⟩ in terms of the basis states |±⟩ is:
|Ψ2⟩ = e^(-i(ϕ/2 + θ/2))|+⟩ + e^(-i(ϕ/2 + θ/2))|-⟩
PART C:
In this part, we need to understand the relationship between |Ψ2⟩ and the state |n;+⟩, where n is the unit vector pointing in the (θ,ϕ) direction.
The state |n;+⟩ can be written as:
|n;+⟩ = cos(θ/2)|+⟩ + e^(iϕ)sin(θ/2)|-⟩
Comparing this with the expression for |Ψ2⟩ obtained in Part B, we can see that they are related. The coefficients of the basis states |+⟩ and |-⟩ in both expressions are the same, but there might be a phase difference.
Therefore, we can write the relationship as:
|Ψ2⟩ = e^(-i(ϕ/2 + θ/2))|n;+⟩