Download Uncertainty Principle Tutorial part II

Uncertainty Principle Tutorial part II
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In quantum mechanics, for any observable A or B , there is a corresponding Hermitian
operator  or B̂ in the Hilbert space in which the state of the system lies.
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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.
2

1

    [ Aˆ , Bˆ ]  is the generalized uncertainty principle that relates the product of
 2i

2
A
2
B
the variances of observables A or B .
ˆ , Bˆ ]  Aˆ Bˆ  Bˆ Aˆ . If  and B̂
The commutator of two operators  and B̂ is defined by [ A
ˆ , Bˆ ]  0 , we call them compatible operators. Otherwise, Â
commute with each other, i.e., [ A
and B̂ 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: Â and B̂ commute
Suppose Hermitian operators  and B̂ commute with each other, answer questions 1 to 3.
1. If
 is an eigenstate of  with eigenvalue  , is B̂ 
what is the eigenvalue of
an eigenstate of  ? If so,
 corresponding to the state B̂  ? Explain. (Hint:
[ Aˆ , Bˆ ]   0 , so Aˆ Bˆ   Bˆ Aˆ   Bˆ   . Remember that  and B̂ have only
non-degenerate eigenstates.)
2. If
 is an eigenstate of  with eigenvalue  , is  an eigenstate of B̂ ? Explain.
(Hint:
 and B̂ 
are both eigenstates of  with the same eigenvalue
proportional to each other, i.e., B̂   
 . Are they
 where  is a constant? Remember that Â
and B̂ have only non-degenerate eigenstates.)
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3. Consider the following conversation between Andy and Caroline:
 n of the operator  must be an eigenstate
Caroline: Now I understand that any eigenstate
of the operator B̂ if  and B̂ commute with each other. But how can we know that
n
also form a complete set of eigenstates for B̂ ?
Andy: When  and B̂ 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  must be an
eigenstate of the operator B̂ . Thus, we can prove that any eigenstate of the operator B̂ must also
be an eigenstate of  if their eigenvalue spectra are non-degenerate. Therefore, the complete set
of eigenstates
 n are shared by both the compatible operators  and B̂ .
Do you agree with Andy? Explain.
4. Suppose operators  and B̂ commute with each other.
both  and B̂ with eigenvalues
 is a common eigenstate of
 and  respectively ( Â     , B̂     ).
At time t=0, the state of the quantum system is
 . If  and B̂ 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
(corresponding to the operator  ) in the state
 if we measure the observable A
 at t=0?
 after your measurement of A at time t=0. If
(b) Suppose the wavefunction collapses to
you measure the observable B (corresponding to the operator B̂ ) 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?
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(c) If you directly measure the observable B in the state
 at time t=0 without
measuring 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  and B̂ 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: Â and B̂ do not commute
ˆ , Bˆ ]  q is a
5. Suppose operators  and B̂ do not commute with each other and [ A
non-zero constant. If
 is an eigenstate of  with nonzero eigenvalue  , could 
also be an eigenstate of B̂ ? Explain. (Hint: Assume B̂   
ˆ , Bˆ ] 
of [ A
 and calculate the value
ˆ , Bˆ ]  q .)
to show that the assumption is incorrect if [ A
ˆ , Bˆ ]  0 but the commutator
6. Suppose operators  and B̂ do not commute with each ( [ A
can either be a constant value or an operator).
 n also be a complete set of eigenstates of B̂ ? Explain.
with eigenvalues  n . Could
(Hint: For any state  
 n form a complete set of eigenstates of Â
c
n
 n , show that [ Aˆ , Bˆ ]   0 if  n form a complete
n
ˆ , Bˆ ]  0 .)
set of eigenstates for both  and B̂ . But this assumption cannot be true if [ A
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 is an eigenstate of
7. Suppose operators  and B̂ do not commute with each other.
 with eigenvalue 
( Â   
and

is an eigenstate of
B̂ with eigenvalue 
 , B̂     ). 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  ) 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 B̂ ) 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  are
probability of the initial state

collapsing to
then calculate the probability of the state
measured.)
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n
 n . Calculate the
 n first when A is measured and
collapsing to

when B is
(d) If we directly measure the observable B in the state
 at time t=0 without
measuring 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  and B̂ 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 ?
 n . If we measure A first, the
Andy: Suppose the eigenstates of the operator  are
probability of the initial state collapsing to  n
is
B following the measurement of A has a probability
Therefore,

n
the
2

 n
overall
2
probability
  n    n
n
2
. The preceding measurement of
 n
2
of
of obtaining the value  .
obtaining
if we measure the observable

is
A before
n
measuring B in the initial state
 .
 directly, the probability of obtaining  is
Caroline : But if we measure B in the state

2
n 
2
. Since
n
form a complete set of vectors,

n
 n  1 , we know
n
      n  n  . Isn’t this probability  
2
n
of getting  if we measure B after the measurement of A ?
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the same as the overall probability

Andy: No. Because the probability of measuring

2

 
directly in the state

is
2
n
n 
. And the probability of obtaining  if we measure B after
n
the measurement of A is

 n n 
2
. Generally,
n
a
n
2
  an .
n
2
n
Do you agree with Andy? Explain whether the probabilities of obtaining the eigenvalue  / 2 are
ˆ  Sˆ , Bˆ  Sˆ and the initial
the same in the situation where the incompatible operators are A
z
x
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  and B̂ 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  and B̂ 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
n
n 
 1 
2
n
   n n 
2
. That’s why the probability does not
n
change when  and B̂ commute with each other as we have seen in question 4(d) .
Do you agree with Andy? Explain.
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9. Which one of the following equations correctly represents the commutation relation between
the x-component and z-component of the angular momentum operators L̂x and L̂y ?
A. [ Lˆ x , Lˆ y ]  0
B. [ Lˆ x , Lˆ y ]  Lˆ z
C. [ Lˆx , Lˆ y ]  Lˆz
D. [ Lˆ x , Lˆ y ]  iLˆ z
10. Can we measure L̂x and L̂y simultaneously in a given quantum state? Explain.
11. According to your answer to the previous question, can you measure the magnitude and




ˆ
direction of the angular momentum operator L  Lˆ x i  Lˆ y j  Lˆ z k simultaneously in a
given 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).
Do you agree with Andy? Explain.
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(b) Consider the following conversation between Andy and Caroline.
Caroline: But what if the expectation value of the commutator
state
[ Aˆ , Bˆ ] is zero in a particular
 . I know for any operator  and B̂ , the generalized uncertainty principle
2
1 ˆ ˆ 
[ A, B ]  . Does that mean that the operators  and B̂ can be measured
gives    
 2i

2
A
2
B
simultaneously in the state 
when
[ Aˆ , Bˆ ]  0 ?
Andy. No. The uncertainty principle is an inequality involving  A2 B2 . It says that  A2 B2 has a
2
1 ˆ ˆ 
[ A, B ]  . In fact,  A2 B2 is usually larger than
 2i

value larger than or equal to 
2
1 ˆ ˆ 
[ A, B ]  . For example, we know that the commutator of the x-component and the

 2i

y-component of the angular momentum operator L̂x and L̂y is [ Lˆx , Lˆ y ]  iLˆz . It is possible
that the expectation value of [ Lˆ x , Lˆ y ] equals zero if the state of the system is 
x
or 
y
.
But since L̂x and L̂y do not commute with each other, i.e., [ Lˆ x , Lˆ y ]  0 , you cannot measure
the x-component and the y-component of the angular momentum simultaneously in a given
quantum state.
Do you agree with Andy? Explain.
Case III: Operators that commute but have some degenerate eigenstates
The energy of a hydrogen atom is En 
 13.6eV
(n=1, 2, 3, …) when we only consider the
n2
Coulomb interaction. The angular momentum eigenstates for the electron in a Hydrogen atom
with energy En can be represented by
, m where   0,1,2,..., n  1 is the quantum
number for the square of the angular momentum operator L̂ and m  ,  1,...,   1,  is
2
the quantum number for the z-component of angular momentum L̂ z . Answer the following
questions.
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13. Which one of the following equations correctly represents the commutation relation between
2
the operators L̂
and L̂ z ?
2
A. [ Lˆ , Lˆ z ]  0
2
B. [ Lˆ , Lˆ z ]  
C. [ Lˆ , Lˆ z ]  i
2
D. [ Lˆ , Lˆz ]  Lx  iL y
2
14. Consider the following conversation between Andy and Caroline.
2
Caroline: Since the square of the angular momentum operator L̂
and the z-component of the
angular momentum operator L̂ z commute with each other, we can infer the value of Lz after
2
the measurement of L , right?
2
Andy: No. L̂
has degenerate eigenvalues (  1)
2
where  is the quantum number
corresponding to the square of the angular momentum. A given value of  corresponds to
2  1 eigenstates , m where m  ,  1,...,   1,  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 ground state of the Hydrogen atom is  
2
you measure L in the state

and obtain 2
2
1
 0,0  1,0  1,1  1,1  . If
2
which corresponds to the quantum number
  1 , the angular momentum state of the electron would collapse to a normalized superposition
of 1,0 and 1,1 , i.e.,
 
1
1,0  1,1  1,1  , which is an eigenstate of L̂2 .
3
However, in this state   , the z-component of angular momentum Lz does not have a definite
value so it is not an eigenstates of L̂ z .
2
Caroline: But what if the measurement of L in the state

collapses the wavefunction to
0,0 ? Can we infer the value of both L2 and Lz in this case?
Andy: Yes. If the state of the system collapse to 0,0
z-component of the angular momentum will have a definite value.
Do you agree with Andy? Explain.
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2
after the measurement of L , the
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
2
square of the angular momentum L after your measurement? Explain.
L̂2 and L̂z (because they
16. We can find a complete set of simultaneous eigenstates for
2
commute although the eigenvalue spectra of L̂
has a degeneracy).
2
Caroline: Now I can see that we may or may not infer the value of L after the measurement of
Lz . Whether L2 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
L̂2 and L̂z commute with each other and they share a complete set of eigenstates , m . It is
true that some superposition of the basis vectors , m
can be an eigenstate of L̂ z but not of
L̂2 , and some other superposition of , m can be an eigenstate of L̂2 but not of L̂ z .
2
However, the probability of obtaining a particular eigenvalue of the operator L̂
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does not depend
on whether we measure Lz first, and vice versa. For example, if we measure L2 directly for the
1
 0,0  1,0  1,1  1,1  , the probability of obtaining   0 is 1 . If we
2
4
1 1
measure Lz first, the probabilities of obtaining m  0 , m  1 and m  1are
,
and
2 4
state  
0,0  1,0
1
respectively. Only when the state of the system collapses to
4
2
after the
measurement of Lz , i.e., m  0 , can we obtain the value   0 if we measure L2 following
the measurement of Lz . Therefore, if we measure Lz first for the state 
L2 , the probability of obtaining   0 is
directly for the state
and then measure
1 1 1
  , which is the same as measuring L2
2 2 4
 .
Do you agree with Andy? Explain.
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