Uncertainty Principle Tutorial Part II

Uncertainty Principle Tutorial part II  In quantum mechanics, for any observable A or B , there is a corresponding Hermitian

operator Aˆ or Bˆ in the Hilbert space in which the state of the system lies.  If two operators are incompatible, measuring an observable corresponding to one operator will affect the probability of measuring the observable corresponding to the other operator.  We can find a complete set of simultaneous eigenstates for compatible operators.  We cannot find a complete set of simultaneous eigenstates for incompatible operators.

1 2 2 2  ˆ ˆ    A B   [A, B]  is the generalized uncertainty principle that relates the product of  2i  the variances of observables A or B .

The commutator of two operators Aˆ and Bˆ is defined by [Aˆ, Bˆ]  AˆBˆ  BˆAˆ . If Aˆ and Bˆ commute with each other, i.e., [Aˆ, Bˆ]  0 , we call them compatible operators. Otherwise, Aˆ and

Bˆ are called incompatible operators. Assume all the operators in this tutorial correspond to physical observables and have non-degenerate eigenstates unless the degeneracy is explicitly mentioned.

Case I: Aˆ and Bˆ commute

Suppose Hermitian operators Aˆ and Bˆ commute with each other, answer questions 1 to 3.

1. If  is an eigenstate of Aˆ with eigenvalue  , is Bˆ  an eigenstate of Aˆ ? If so, what is

the eigenvalue of Aˆ corresponding to the state Bˆ  ? Explain. (Hint: [Aˆ, Bˆ]  0 , so

AˆBˆ   BˆAˆ   Bˆ  . Remember that Aˆ and Bˆ have only non-degenerate eigenstates.)

2. If  is an eigenstate of Aˆ with eigenvalue  , is  an eigenstate of Bˆ ? Explain. (Hint:

 and Bˆ  are both eigenstates of Aˆ with the same eigenvalue  . Are they proportional

to each other, i.e., Bˆ     where  is a constant? Remember that Aˆ and Bˆ have only non-degenerate eigenstates.)

1 3. Consider the following conversation between Andy and Caroline:

Caroline: Now I understand that any eigenstate  n of the operator Aˆ must be an eigenstate of the operator Bˆ if Aˆ and Bˆ commute with each other. But how can we know that  n also form a complete set of eigenstates for Bˆ ?

Andy: When Aˆ and Bˆ have only non-degenerate eigenstates, by following the same method as in the previous question we can show that any eigenstate  n of the operator Aˆ must be an eigenstate of the operator Bˆ . Thus, we can prove that any eigenstate of the operator Bˆ must also be an eigenstate of Aˆ if their eigenvalue spectra are non-degenerate. Therefore, the complete set of eigenstates  n are shared by both the compatible operators Aˆ and Bˆ . Do you agree with Andy? Explain.

4. Suppose operators Aˆ and Bˆ commute with each other.  is a common eigenstate of both

Aˆ and Bˆ with eigenvalues  and  respectively ( Aˆ     , Bˆ     ). At time

t=0, the state of the quantum system is  . If Aˆ and Bˆ have only non-degenerate eigenstates, answer the following questions and express the probabilities in Dirac notation. (a) What is the probability of obtaining the value  if we measure the observable A

(corresponding to the operator Aˆ ) in the state  at t=0?

(b) Suppose the wavefunction collapses to  after your measurement of A at time t=0. If

you measure the observable B (corresponding to the operator Bˆ ) immediately after the measurement of A at time t=0, what value will you get? What is the overall probability

(compared to the initial state  ) of obtaining this value?

(c) If you directly measure the observable B in the state  at time t=0 without measuring

2 A first, what is the probability of obtaining the value  ? Is this result the same as your

answer in the previous problem (question b)? Explain.

(d) When Aˆ and Bˆ are compatible operators with non-degenerate eigenstates and the initial

state of the system is  , does the probability of obtaining  depend on whether we

measure A first? Explain.

Case II: Aˆ and Bˆ do not commute

5. Suppose operators Aˆ and Bˆ do not commute with each other and [Aˆ, Bˆ]  q is a non-zero

constant. If  is an eigenstate of Aˆ with nonzero eigenvalue  , could  also be an

eigenstate of Bˆ ? Explain. (Hint: Assume Bˆ     and calculate the value of [Aˆ, Bˆ]

to show that the assumption is incorrect if [Aˆ, Bˆ]  q .)

6. Suppose operators Aˆ and Bˆ do not commute with each ([Aˆ, Bˆ]  0 but the commutator can

either be a constant value or an operator).  n form a complete set of eigenstates of Aˆ with

eigenvalues n . Could  n also be a complete set of eigenstates of Bˆ ? Explain. (Hint: For

  c  ˆ ˆ any state  n n , show that[A, B]   0 if  n form a complete set of n

eigenstates for both Aˆ and Bˆ . But this assumption cannot be true if [Aˆ, Bˆ]  0 .)

7. Suppose operators Aˆ and Bˆ do not commute with each other.  is an eigenstate of Aˆ

3 with eigenvalue  and  is an eigenstate of Bˆ with eigenvalue  ( Aˆ     ,

Bˆ     ). At time t=0, the state of the quantum system is  . Answer the following questions and express the probability in Dirac notation.

(a) At time t=0, if we measure the observable A (corresponding to the operator Aˆ ) in the

state  , what is the probability of obtaining the value  ? Explain.

(b) Suppose the measurement of A collapses the wavefunction into the state  . If we

measure the observable B (corresponding to the operator Bˆ ) immediately after the

measurement of A at time t=0, what is the probability (compared to the state  ) of

obtaining the value  ? Explain.

(c) In the previous problem (question 7.b), if we don’t know the outcome after the

measurement of A , what is the overall probability (compared to the initial state  ) of

obtaining the value  if we measure B after the measurement of A ? Explain. (Hint:

suppose the complete set of eigenstates of the operator Aˆ are  n . Calculate the

probability of the initial state  collapsing to  n first when A is measured and then

calculate the probability of the state  n collapsing to  when B is measured.)

(d) If we directly measure the observable B in the state  at time t=0 without measuring

4 A first, what is the probability of obtaining the value  ? Is this result the same as your

answer in the previous problem (question c)? Explain.

(e) When Aˆ and Bˆ are incompatible operators with non-degenerate eigenstates and the

initial state of the system is  , does the probability of obtaining  depend on whether

we measure A first? Explain.

8. (a) Consider the following conversation between Andy and Caroline:

Caroline: I don’t understand the answer to the previous problem. How can the measurement of A affect the overall probability of obtaining the value  of the observable B ?

Andy: Suppose the eigenstates of the operator Aˆ are  n . If we measure A first, the

2 probability of the initial state collapsing to  n is  n  . The preceding measurement of B

2 following the measurement of A has a probability   n of obtaining the value  .

Therefore, the overall probability of obtaining  is

2 2 2   n    n   n    n if we measure the observable A before measuring n n

B in the initial state  .

Caroline : But if we measure B in the state  directly, the probability of obtaining  is

2   1   . Since  n form a complete set of vectors,  n n , we know n

2       n  n  . Isn’t this probability   the same as the overall probability n of getting  if we measure B after the measurement of A ?

Andy: No. Because the probability of measuring  directly in the state  is

5 2 2 . And the probability of obtaining  if we measure after the       n  n  B n

2 2 2 measurement of is   n  n  . Generally, . A  an   an n n n Do you agree with Andy? Explain whether the probabilities of obtaining the eigenvalue / 2 are ˆ ˆ ˆ ˆ the same in the situation where the incompatible operators are A  Sx , B  Sz and the initial

state is   z for the following cases.

(i) B is measured directly in the state   z .

(ii) B is measured after the measurement of A in the state   z .

(b) Consider the following conversation between Andy and Caroline: Caroline: Now I understand that when the operators do not commute, the probability of measuring B depends on whether A is measured before in the state  . However, if the operators Aˆ and Bˆ commute with each other, why is the probability of measuring B the same in the state  whether it is measured directly or after the measurement of A ?

Andy: When the Hermitian operators Aˆ and Bˆ are compatible, they have a complete set of

simultaneous eigenstates  n . Let’s consider the probability of obtaining a particular eigenvalue

 when measuring the observable B . Suppose the eigenstate corresponding to the eigenvalue 

is   1 . We know   n equals zero for all the  n except for n  1. So

2 2 2 . That’s why the probability does not    n  n    1      n  n  n n change when Aˆ and Bˆ commute with each other as we have seen in question d .

Do you agree with Andy? Explain.

9. Which one of the following equations correctly represents the commutation relation between

6 ˆ ˆ the x-component and z-component of the angular momentum operators Lx and Ly ?

ˆ ˆ A. [Lx , Ly ]  0

ˆ ˆ ˆ B. [Lx , Ly ]  Lz

ˆ ˆ ˆ C. [Lx , Ly ]  Lz

ˆ ˆ ˆ D. [Lx , Ly ]  iLz

ˆ ˆ 10. Can we measure Lx and Ly simultaneously in a given quantum state? Explain.

11. According to your answer to the previous question, can you measure the magnitude and     direction of the angular momentum operator ˆ ˆ ˆ ˆ simultaneously in a given L  Lxi  Ly j  Lzk quantum state? Explain.

12. (a) Consider the following conversation between Andy and Caroline. Caroline: Does the state of the system determine whether you can measure two observables simultaneously? Andy: No. Whether you can measure two observables simultaneously only depends on the commutation relation between the corresponding operators. Only when the operators are compatible can we measure the corresponding observables at the same time (both observables can be well-defined in a given quantum state).

(b) Consider the following conversation between Andy and Caroline.

7 Caroline: But what if the expectation value of the commutator [Aˆ, Bˆ] is zero in a particular state  . I know for any operator Aˆ and Bˆ , the generalized uncertainty principle gives

2 2 2  1 ˆ ˆ  ˆ ˆ  A B   [A, B]  . Does that mean that the operators A and B can be measured  2i  simultaneously in the state  when [Aˆ, Bˆ]  0 ?

2 2 2 2 Andy. No. The uncertainty principle is an inequality involving  A B . It says that  A B has a

2  1 ˆ ˆ  2 2 value larger than or equal to  [A, B]  . In fact,  A B is usually larger than  2i 

2  1   [Aˆ, Bˆ]  . For example, we know that the commutator of the x-component and the y-  2i 

ˆ ˆ ˆ ˆ ˆ component of the angular momentum operator Lx and Ly is [Lx , Ly ]  iLz . It is possible that the expectation value of [Lˆ , Lˆ ] equals zero if the state of the system is  or  . But x y x y

ˆ ˆ ˆ ˆ since Lx and Ly do not commute with each other, i.e., [Lx , Ly ]  0 , you cannot measure the x- component and the y-component of the angular momentum simultaneously in a given quantum state.

Case III: Operators that commute but have some degenerate eigenstates 13.6eV The energy of a hydrogen atom is E  (n=1, 2, 3, …) when we only consider the n n2 Coulomb interaction. The angular momentum eigenstates for the electron in a Hydrogen atom with energy En can be represented by ℓ,m where ℓ  0,1,2,...,n 1 is the quantum number for the square of the angular momentum operator Lˆ2 and m  ℓ,ℓ1,...,ℓ1,ℓ is the

ˆ quantum number for the z-component of angular momentum Lz . Answer the following questions.

13. Which one of the following equations correctly represents the commutation relation between

8 ˆ2 ˆ the operators L and Lz ?

ˆ2 ˆ A. [L , Lz ]  0

ˆ2 ˆ B. [L , Lz ]  

ˆ2 ˆ C. [L , Lz ]  i

ˆ2 ˆ D. [L , Lz ]  Lx  iLy

14. Consider the following conversation between Andy and Caroline.

Caroline: Since the square of the angular momentum operator Lˆ2 and the z-component of the

ˆ angular momentum operator Lz commute with each other, we can infer the value of Lz after the measurement of L2 , right? Andy: No. Lˆ2 has degenerate eigenvalues ℓ(ℓ1)2 where ℓ is the quantum number corresponding to the square of the angular momentum. A given value of ℓ corresponds to 2ℓ1 eigenstates ℓ,m where m  ℓ,ℓ1,...,ℓ1,ℓ is the quantum number for the z-component of angular momentum. For example, consider the case when the initial state of the electron in the 1 ground state of the Hydrogen atom is    0,0  1,0  1,1  1,1 . If you measure L2 in 2 the state  and obtain 22 which corresponds to the quantum number ℓ  1, the angular momentum state of the electron would collapse to a normalized superposition of 1,0 and 1,1 ,

1 i.e.,    1,0  1,1  1,1 , which is an eigenstate of Lˆ2 . However, in this state   , 3

the z-component of angular momentum Lz does not have a definite value so it is not an

ˆ eigenstates of Lz .

Caroline: But what if the measurement of L2 in the state  collapses the wavefunction to

2 0,0 ? Can we infer the value of both L and Lz in this case?

Andy: Yes. If the state of the system collapse to 0,0 after the measurement of L2 , the z- component of the angular momentum will have a definite value.

9 15. Suppose the initial state of the electron in the ground state of the Hydrogen atom is 1    0,0  1,0  1,1  1,1 . Use the previous question (question 14) as a guide to 2 answer questions 15(a) and 15(b).

(a) If you measure the z-component of the angular momentum Lz in the state  and obtain the eigenvalue 0 (corresponding to the quantum number m  0 ), does the electron have a definite value of the square of angular momentum L2 after your measurement? Explain.

(b) If you measure the z-component of the angular momentum Lz for the state  and obtain  (corresponding to the quantum number m  1), does the electron have a definite value of the square of the angular momentum L2 after your measurement? Explain.

ˆ2 ˆ 16. We can find a complete set of simultaneous eigenstates for L and Lz (because they

commute although the eigenvalue spectra of Lˆ2 has a degeneracy). Caroline: Now I can see that we may or may not infer the value of L2 after the measurement of

2 Lz . Whether L has a definite value after the measurement of Lz depends on what state the

2 system collapses into. But does it mean that we cannot measure L and Lz simultaneously in any state?

2 Andy: No. The observables L and Lz can be measured simultaneously because the operators

ˆ2 ˆ ℓ L and Lz commute with each other and they share a complete set of eigenstates ,m . It is true

ℓ ˆ ˆ2 that some superposition of the basis vectors ,m can be an eigenstate of Lz but not of L , and

ℓ ˆ2 ˆ some other superposition of ,m can be an eigenstate of L but not of Lz . However, the

10 probability of obtaining a particular eigenvalue of the operator Lˆ2 does not depend on whether we

2 measure Lz first, and vice versa. For example, if we measure L directly for the state 1 1    0,0  1,0  1,1  1,1 , the probability of obtaining ℓ  0 is . If we measure 2 4 1 1 1 L first, the probabilities of obtaining m  0 , m  1 and m  1are , and z 2 4 4 0,0  1,0 respectively. Only when the state of the system collapses to after the measurement 2

2 of Lz , i.e., m  0 , can we obtain the value ℓ  0 if we measure L following the measurement

2 of Lz . Therefore, if we measure Lz first for the state  and then measure L , the probability of 1 1 1 obtaining ℓ  0 is   , which is the same as measuring L2 directly for the state  . 2 2 4