Larmor Precession Tutorial

Here is instructions on how to use the simulation program.(The first simulation is used in question 5) 1. Double click the file “osp_spins” 2. On the top of the new window, click “FileÆOpen Internal” 3. Choose the second file “osp_spins_about.xset” 4. On the left column of the new window, double click “OSP Spins Program” 5. Double click “custom configuratin” 6. In the new window, click “FileÆLoad XML” , and then choose an xml file to open the corresponding simulation.

Initial state ↑ , measure Sˆ (Question 1~7) z z

r ˆ ˆ 1. The Hamiltonian of a spin-1/2 particle in a magnetic field = 0 zBB ˆ is −= γ 0SBH z , where

ˆ h ⎛ 01 ⎞ Sz = ⎜ ⎟ is the z-component of spin angular momentum. Which of the following is 2 ⎝ −10 ⎠

ˆ ˆ the eigenvalue of the Hamiltonian −= γ 0SBH z ?

A. n = mγBE 0 γB B. E = 0 n m 2

C. = h γBE n m 2 0

D. E = h n m 2

r 2. An electron in a magnetic field = zBB ˆ is initially in the state χ )0( ↑= . Which of 0 z

the following equation correctly represents the state χ t)( of the electron? The

ˆ ˆ Hamiltonian operator is −= γ 0SBH z .

A. χ t)( ↑= z

B. χ )( et γ 0tBi 2/ ↑⋅= z

C. χ )( et γ 0tBi 2/ e− γ 0tBi 2/ ↓⋅+↑⋅= z z

D. χ )( aet γ 0tBi 2/ be− γ 0tBi 2/ ↓⋅+↑⋅= z z 3. Consider the following conversation between Andy and Caroline about the above Hamiltonian operator. ˆ ˆ Andy: H is essentially Sz except for some multiplicative constants. Therefore, the

ˆ ˆ eigenstates of Sz will also be the eigenstates of H .

ˆ ˆ Caroline: No. The presence of magnetic field will make the eigenstates of Sz and H different. The eigenstates of Hˆ will change with time in a non-trivial manner.

ˆ Andy: I disagree. If the magnetic field had a time dependence, e.g., = 0 ω )cos( ktBB , the eigenstates of Hˆ will change with time in a non-trivial manner but not for the present

r case where B is constant.

With whom do you agree? A. Andy B. Caroline C. Neither

4. In the previous problem (problem 2), suppose t0 satisfies − γ tB 00 = π 6/2/ . What is the

probability of getting ↑ when we measure Sˆ at = tt ? z z 0

5. Now you can use “FileÆLoad XMLÆ larmor_z-up_Sz_30” to check your answer to the previous problem (problem 4). In the simulation, the red block is the magnetic ˆ field, and the blue block is the Stern-Gerlach apparatus to measure Sz . When the

upper detector after SGZ clicks, the state of the particle measured is ↑ . The z number “30” in the magnetic field means that the particle will stay in the magnetic field

o for time t0 satisfying tB 00 πγ ==− 306/2/ . Explain whether what you observed in the

simulation is consistent with your prediction in the previous question. If it is not consistent, resolve this discrepancy.

6. In the previous problem (problem 2), does the probability of getting ↑ depend on the z time t ? Explain.

Now you can use the simulations “larmor_z-up_Sz_30”, “larmor_z-up_Sz_45” and

“larmor_z-up_Sz_90” to check your answer. In these three simulation, the time t0

o o o satisfies tB 00 πγ ==− 306/2/ , tB 00 πγ ==− 454/2/ and tB 00 πγ ==− 902/2/ respectively.

7. Explain whether what you observed in the simulation above is consistent with your prediction in the previous question. If it is not consistent, resolve this discrepancy.

Initial state ↑ , measure Sˆ (Question 8~15) z x r 8. An electron in a magnetic field = zBB ˆ is initially in the state χ )0( ↑= . After time t, 0 z

the electron will be in state χ t)( . How can you express χ t)( in the spin-x basis ↑ x

and ↓ ? x

9. In the previous problem (problem 8), suppose t0 satisfies that − γ tB 00 = π 6/2/ , what is

the probability of getting ↑ when we measure the observable S at = tt ? x x 0

10. Now you can use the simulations “larmor_z-up_Sx_30” to check your answer. Explain whether what you observed in the simulation is consistent with your prediction in the previous question. If it is not consistent, resolve this discrepancy.

11. In the previous problem (problem 8), does the probability of getting ↑ depend on the x time t ?

Now you can use the simulations “larmor_z-up_Sx_30” , “larmor_z-up_Sx_45” and

“larmor_z-up_Sx_90” to check your answer. In these three simulation, the time t0

o o o satisfies tB 00 πγ ==− 306/2/ , tB 00 πγ ==− 454/2/ and tB 00 πγ ==− 902/2/ respectively.

12. Explain whether what you observed in the simulation above is consistent with your prediction in the previous question. If it is not consistent, resolve this discrepancy.

ˆ ˆ 13. Calculate the expectation value of Sx at time t . Then evaluate Sx at time t0

o satisfying tB 00 πγ ==− 306/2/ .

14. Find the expectation value by counting the number of clicks in the upper and lower detectors in the simulation “larmor_z-up_Sx_30”? Is this simulation result the same as what you calculated in the previous question (problem 13)?

15. If you had to explain to a friend what expectation value physically means and why the calculated value and the value from the simulation are the same, what would you say?

Initial state χ )0( ba ↓+↑= , measure Sˆ (Question 16~23) z z z

16. If the state of the system at an initial time t = 0 is given by χ )0( ba ↓+↑= , z z

which one of the following is the state, χ t)( , after a time t? The Hamiltonian operator is

ˆ ˆ −= γ 0SBH z .

A. χ )0( γ 0tBi 2/ ( bae ↓+↑⋅= ) z z

B. χ )0( − γ 0tBi 2/ ( bae ↓+↑⋅= ) z z

C. χ )0( γ 0tBi 2/ ( babae )()( ↓−+↑+⋅= ) z z

D. χ )0( ae γ 0tBi 2/ be− γ 0tBi 2/ ↓⋅+↑⋅= z z

ˆ 17. In problem 16, which one of the following gives the correct outcomes when we measure Sz

ˆ ˆ at = tt 0 ? The Hamiltonian operator is −= γ 0SBH z .

γ tBi 2/ − γ tBi 2/ A. h 2/ with a probability ae 0 and − h 2/ with a probability be 0

2 γ tBi 2/ 2 − γ tBi 2/ B. h 2/ with a probability ea 0 and − h 2/ with a probability eb 0

2 2 C. h 2/ with a probability a and − h 2/ with a probability b

D. h 2/ and − h 2/ with equal probability

ˆ 18. Consider the following conversation between Andy and Caroline about measuring Sz in

the state χ t)(

2 2 Andy: Since the probability of measuring h 2/ is a , − h 2/ is b and

ba 22 =+ 1, we can choose our a and b as = ea iφ1 α)cos( and = eb iφ2 α)sin(

where α , φ1 and φ2 are free parameters.

Caroline: I agree. Since 2 2 αα =+ 1)(sin)(cos , it gives the same relation as

ba 22 =+ 1and there is no loss of generality.

1 Do you agree with Andy and Caroline? If yes, for the state χ )0( ( ↓+↑= ), what 2 zz are the values of a and b when you express it in terms of χ )0( ba ↓+↑= ? And what z z

are the values of α , φ1 and φ2 when you express it in terms of α defined earlier as

= ea iφ1 α)cos( and b = eiφ2 α)sin( .

19. Suppose the initial state is χ cos)0( α sinα ↓+↑= , which one of the following is z z ˆ ˆ ˆ ˆ the expectation value z = )( z χχ tStS )( ? The Hamiltonian operator is −= γ 0SBH z .

A. α γ 0tB h 2/)cos()2cos(

B. α γ 0tB h 2/)sin()2sin(

C. α h 2/)2cos(

D. α h 2/)2sin(

Now you can check your answers of the previous problems with a concrete example below (problem 20 to 23). r 1 20. An electron in a magnetic field = 0 zBB ˆ is initially in the state χ )0( ()↓+↑= . 2 zz

Suppose t satisfies that − γ tB = π 6/2/ . What is the probability of getting ↑ when 0 00 z

ˆ ˆ we measure the observable Sz at = tt 0 ? The Hamiltonian operator is −= γ 0SBH z .

Now you can use “FileÆLoad XMLÆ larmor_z-up+z-down_Sz_30” to check your answer.

21. Explain whether what you observed in the simulation above is consistent with your prediction in the previous question. If it is not consistent, resolve this discrepancy.

22. In problem 21, does the probability of getting ↑ depend on the time t ? What is the z

ˆ expectation value of Sz ?

Now you can use “larmor_z-up+z-down_Sz_30”, “larmor_z-up+z-down_Sz_45” and “larmor_z-up+z-down_Sz_90” to check your answer. In these three simulation, the time

o o o t0 satisfies tB 00 πγ ==− 306/2/ , tB 00 πγ ==− 454/2/ and tB 00 πγ ==− 902/2/ respectively.

23. Explain whether what you observed in the simulation above is consistent with your prediction in the previous question. If it is not consistent, resolve this discrepancy.

Initial state χ )0( ba ↓+↑= , measure Sˆ (Question 24~29) z z x

Now let’s go back to the most general initial state at time t = 0 which is given by

χ )0( ba ↓+↑= . The Hamiltonian is ˆ −= γ SBH ˆ . Answer problem 26 and 27. z z 0 z

24. Find the state χ t)( at time t in the spin-x basis ↑ and ↓ . Express your answer x x

in terms of α defined earlier as = ea iφ1 α)cos( and = eb iφ2 α)cos( .

25. Suppose the initial state is χ cos)0( α sinα ↓+↑= , which one of the following is z z ˆ ˆ ˆ ˆ the expectation value x = )( x χχ tStS )( ? The Hamiltonian operator is −= γ 0SBH z .

A. α γ 0tB h 2/)cos()2cos(

B. α γ 0tB h 2/)cos()2sin(

C. α h 2/)2cos(

D. α h 2/)2sin( Now you can check your answers of the previous problems with a concrete example in which

φ1 = 0 , φ2 = 0 , α = π 4/ (problem 26 to 29).

r 1 26. An electron in a magnetic field = 0 zBB ˆ is initially in the state χ )0( ()↓+↑= . 2 zz

The Hamiltonian operator is ˆ −= γ SBH ˆ . What is the probability of getting ↑ at time t? 0 z x

Does the probability of getting ↑ depend on the time t ? Explain. x

Now you can use “larmor_z-up+z-down_Sx_30” , “larmor_z-up+z-down_Sx_45” and “larmor_z-up+z-down_Sx_90”to check your answer. In these three simulation, the time

o o o t0 satisfies tB 00 πγ ==− 306/2/ , tB 00 πγ ==− 454/2/ and tB 00 πγ ==− 902/2/ respectively.

27. In problem 26, the initial state is just χ )0( ↑= , but why the probability of getting ↑ x x

is time-dependent? Can we write the state χ t)( at time t as χ )( et γ 0tBi 2/ ↑⋅= ? x Explain.

ˆ ˆ 28. In problem 26, what is the expectation value of Sx at time t ? Does Sx depend on time?

ˆ When t0 satisfies that − γ tB 00 = π 6/2/ , what is Sx at = tt 0 ?

Now you can use “larmor_z-up+z-down_Sx_30 ” to check your answer. 29. Explain whether what you observed in the simulation above is consistent with your prediction in the previous question.

Initial state χ )0( ba ↓+↑= , measure Sˆ and find the general principal of z z y time-dependence in Larmor precession.(Question 30~39)

r 1 30. An electron in a magnetic field = 0 zBB ˆ is initially in the state χ )0( ()↓+↑= . 2 zz

The Hamiltonian operator is ˆ −= γ SBH ˆ . What is the probability of getting ↑ if we 0 z y

ˆ measure S y at = tt 2 ? t2 satisfies that −γ tB 20 = π 4/2/

Now you can use “FileÆLoad XMLÆ larmor_z-up+z-down_Sy ” to check your answer. 31. Explain whether what you observed in the simulation above is consistent with your prediction in the previous question.

ˆ ˆ 32. Explain in your own words why Sz above does not depend on time where as Sx

ˆ and S y do.

ˆ 33. Consider the following conversation between Pria and Mira about Sz in state χ t)( :

ˆ Pria: Since the state of the system χ t)( evolves in time, the expectation value Sz will depend on time. Mira: I disagree with the second part of your statement. The time development of the ˆ Ad i ∂Aˆ expectation value of any operator Aˆ is given by = []ˆ , AH ˆ + . In our case, dt h ∂t ˆ ˆ ˆ ˆ ∂Sz [ ,SH z ]= 0 and the operator Sz does not have any explicit time dependence so = 0 . ∂t

ˆ Sd z Thus, = 0 and the expectation value will not change with time. dt With whom do you agree?

ˆ ˆ In problem 34~39, the Hamiltonian operator is −= γ 0SBH z . r 34. Choose all of the following statements that are true about the expectation value S in

the state χ t)( :

r A. S will always depend on time because [ ˆ ,SH ˆ ≠ 0]

r B. S cannot depend on time because the expectation value of an observable is its

time-averaged value. r C. S is time-independent only when the initial state is purely ↑ or ↓ z z

r 35. Choose all of the following statements that are true about the expectation value S in

the state χ t)( when the initial state is not purely ↑ or ↓ : z z r ˆ (1) The z component of S , i.e., Sz , is time-independent.

r ˆ ˆ (2) The x and y components of S change with time. When Sx is maximum S y

is a minimum and vice versa. ˆ ˆ (3) The magnitude of the maximum values of Sx and S y are the same.

A. 1 and 2 B. 1 and 3 C. 2 and 3 D. All of the above

r 36. Choose all of the following statements that are true about the expectation value S in

the state χ t)( :

r (1) The vector S can be thought to be precessing about the z axis at an angle 2α .

r (2) The vector S can be thought to be precessing about the z axis with a

frequencyω = γB0 . r (3) All the three components of vector S change as it precesses about the z axis.

A. 1 and 2 B. 1 and 3 C. 2 and 3 D. All of the above

37. Choose a three dimensional coordinate system in the spin space with the z axis in the r vertical direction. Draw a sketch showing the precession of S about the z axis when

r the state of the system starts out in χ )0( ba ↓+↑= . Show the angle that S z z makes with the z axis and the precession frequency explicitly. r Show the projection of S along the x, y and z axes at two separate times. Explain in

r words why the projection of S along the z direction does not change with time but

those along the x and y directions change with time.

38. Explain in your own words, why for ˆ −= γ SBH ˆ , if the initial state is ↑ , Sˆ , 0 z z x

ˆ ˆ 1 S y and Sz are all time independent. But if the initial state is ( ↓+↑ ) , then 2 zz

ˆ ˆ ˆ Sx and S y depend on time while Sz does not. Your explanation should bring out

general principle of when expectation value of an operator will not depend on time.