How Does an Extra Factor of i Affect Quantum Probability Calculations?

[davon806]

Homework Statement

I am not sure about (c) and (d). Firstly, I calculated the eigenvector of A :
|v_1> = ( |2 > - |1> )/ √(2) ,eigenvalue -2
|v_2> = ( |2> + |1>) / √(2) , eigenvalue 2

For (c), basically it follows from part (b) where the probability of a_1 is given by the formula | <v_1 | ψ > |^2 , and similarly for a_2 ( using v_2)

However, I don't see any reason that an extra factor of i will change the value of those probabilities ? ( I've calculated a_1 and a_2 which are both 1/2, in (b) )

For(d), I would like to use the formula ψ(t) = e^(-iEt/h) ψ(0). However, the eigenstates of the Hamiltonian was not provided. Given the fact that we only have |1> , |2> , |v_1> and |v_2> : How can I use this formula?

Homework Equations

In the general case, the Hamiltonian of the system can be written as a 2 × 2 matrix, where the elements of the matrix are given by: H_ij = < i | H | j > . So for example, <2 | A | 1 > = A_21 = 2. Actually A is not the Hamiltonian so I am not even sure whether it is applicable to part (a). Nevertheless I use this in (a). If I was wrong please correct me !

The Attempt at a Solution

Incorporated in question

 

[TSny]

Homework Statement

Firstly, I calculated the eigenvector of A :
|v_1> = ( |2 > - |1> )/ √(2) ,eigenvalue -2
|v_2> = ( |2> + |1>) / √(2) , eigenvalue 2

OK

For (c), basically it follows from part (b) where the probability of a_1 is given by the formula | <v_1 | ψ > |^2 , and similarly for a_2 ( using v_2) However, I don't see any reason that an extra factor of i will change the value of those probabilities ? ( I've calculated a_1 and a_2 which are both 1/2, in (b) )

The factor of $i$ does make a difference. Did you try calculating the probabilities for this case?

For(d), I would like to use the formula ψ(t) = e^(-iEt/h) ψ(0). However, the eigenstates of the Hamiltonian was not provided.

Actually, the eigenstates of H are essentially given. Note that the form of the matrix given for H is for the original basis {|1>, |2>}. What do you get for H|1>?

Actually A is not the Hamiltonian so I am not even sure whether it is applicable to part (a). Nevertheless I use this in (a). If I was wrong please correct me !

You are OK here.

 

[davon806]

OK

The factor of $i$ does make a difference. Did you try calculating the probabilities for this case?

Actually, the eigenstates of H are essentially given. Note that the form of the matrix given for H is for the original basis {|1>, |2>}. What do you get for H|1>?
You are OK here.

Hi TSny ,

For (b), firstly I rewrite | ψ > = (1/2√2) ( |v_2> - |v_1> + i√3 |v_1> + i√3 |v_2>)
then I calculated P_a1 = | < v_1 | Ψ > | ^2 = (1/8) | -1 + √3 i |^2 = (1/8) *4 = 1/2 .

Hence we can calculate Φ in a similar way : | Φ > = (1/2√2) ( |v_2> - |v_1> + √3 |v_1> + √3 |v_2>) , so
new P_a1 = | < v_1 | Φ > | ^2 = (1/8) (-1 + √3 )^2 = (1/8) *(2+2√3 ) = (1+√3 )/ 4 > old P_a1?

What is the physical reason behind this? (As asked in (c))

For(d), write |1> = (x y) . H |1> = (E1 0 E2 0) ( x y) = (E1x E2y) ≠ λ|1> ?

 

[TSny]

For (b), firstly I rewrite | ψ > = (1/2√2) ( |v_2> - |v_1> + i√3 |v_1> + i√3 |v_2>)
then I calculated P_a1 = | < v_1 | Ψ > | ^2 = (1/8) | -1 + √3 i |^2 = (1/8) *4 = 1/2 .

OK

Hence we can calculate Φ in a similar way : | Φ > = (1/2√2) ( |v_2> - |v_1> + √3 |v_1> + √3 |v_2>) , so
new P_a1 = | < v_1 | Φ > | ^2 = (1/8) (-1 + √3 )^2 = (1/8) *(2+2√3 ) = (1+√3 )/ 4 > old P_a1?

OK, except you didn't quite calculate (-1 + √3 )^2 correctly.

What is the physical reason behind this? (As asked in (c))

I don't think I can give you a decent physical reason. But it's important to understand that when you expand a state in terms of basis vectors, the relative phases of the coefficients of the expansion are very important. Changing the relative phase changes the state. So, the state (1/√2)( |1> + |2> ) is a different state than (1/√2)( |1> + i |2> ), even though the magnitudes of the coefficients are the same in each state.

For example, if |1> represents the spin state of a spin 1/2 particle where the spin is definitely "up along the z-direction" and if |2> represents the state of the particle where the spin is definitely "down along the z-direction", then (1/√2)( |1> + |2> ) is the state where the spin is definitely along the +x direction while (1/√2)( |1> + i |2> ) is the state where the spin is definitely along the +y direction. So, the presence of the i makes a big difference.

For(d),

write |1> = (x y) . H |1> = (E1 0 E2 0) ( x y) = (E1x E2y) ≠ λ|1> ?

You are working with matrices that are written with respect to the basis vectors |1> and |2>. So, the column vector that represents |1> would be (1 0)^T.

 

[davon806]

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