Download 8 The Heisenberg`s Uncertainty Principle

Course Number: C668
Quantum Mechanics
8 The Heisenberg’s Uncertainty Principle
1. Simultaneous eigenstates: Consider two operators that commute:
h
i
Â, B̂ = 0
(8.0.1)
Let  satisfy the following eigenvalue equation:
 |ηi = a |ηi
(8.0.2)
B̂ Â |ηi = B̂ [a |ηi] = aB̂ |ηi = a |χi
(8.0.3)
Multiplying both sides by B̂
h
i
where we have assumed B̂ |ηi ≡ |χi.
Now lets try and use the commutation relation:
h
i
h
i
B̂ Â |ηi = Â B̂ |ηi = Â |χi
(8.0.4)
From the right hand sides of the last two equations it follows that |χi is also an eigenvector
of  with eigenvalue a. This could only happen if:
(a) |χi = b |ηi since |ηi is an eigenvector of  with eigenvalue a. Hence commuting
operators have simultaneous eigenstates. That is these can be exactly measured simultaneously. (By extension two operators that do not commute cannot be measured
simultaneously as we will see in the next section. ) This is a very important distinguishing factor with respect to classical mechanics. In classical mechanics you can
measure any two observables simultaneously. In quantum mechanics, only variables
whose (Hermitian) operators commute can be observed simultaneously. Now go back
and read the Stern-Gerlach experiment. It must mean that since Sz , Sx and Sy cannot be
observed simultaneously, the operators corresponding to these observables, the Pauli
spin matrices, do not commute.
(b) There is a second possibility and that is |χi is not equal to a constant times |ηi. In this
case the operator  must have degenerate states (|χi and |ηi) and the operator B̂ can be
used to obtain all the degenerate states of Â. But even in this case, the non-degenerate
eigenvectors of  are simultaneously eigenstates of B̂ as is clear from the previous
case.
2. Actually case (b) is also deeply connected with Stern-Gerlach as we will see later.
3. Expectation or Average value of an operator
If the |ψi represents the state of a system, the expectation value of an operator  with respect
to that state is given by the quantity
D E
 = hψ|  |ψi
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(8.0.5)
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Course Number: C668
Quantum Mechanics
D E
What does this mean? If there is an observable associated with the operator Â, then  is
the average measured value of the observable.
(a) Case 1: The state vector ψ Dis an
E eigenstate of the operator Â. In that case, if a is the
associated eigenvalue then  = hψ|  |ψi = a. (Note: hψ| ψi = 1.) That is the
observable has a measured value equal to the eigenvalue of the operator Â. Consider
Figure 6 (reproduced below) as a way to understand this concept. Here, an Sz+ state is
sent into an SGz measuring device and the resultant measurement is the eigenvalue of
the Sz+ state with respect to the Sˆz operator. Hence if ψ is an eigenstate of the operator,
the corresponding measured value, or expectation value is a,
Figure 19:
(b) Case 2: The state vector ψ is not an eigenstate of the operator Â. In this case, if Â
is a Hermitian operator then the eigenstates of a Hermitian operator form a complete
ortho-normal set. Hence any vector like ψ can be expanded as a linear combination
of such a complete set of vectors. Therefore, if {χi } are the family of eigenvectors of
Â, with eigenvalues {ai }, that is
Âχi = ai χi
i = 1, 2, 3, · · ·
(8.0.6)
then we can expand the vector ψ as a linear combination of the vectors {χi }, as
|ψi =
"
X
#
|χi i hχi | |ψi =
i
X
|χi i hχi | |ψi
(8.0.7)
i
Note we have used the “resolution of identity” here. If we denote the number hχi | |ψi =
ci (note that this is a “dot” product and hence is a number), then
|ψi =
X
ci |χi i
(8.0.8)
i
The expectation value of the operator  can then be written as
D E
 = hψ|  |ψi =
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
X

j


cj ∗ hχj | Â

(
X
i
)
ci |χi i
c
2014,
Srinivasan S. Iyengar (instructor)
Course Number: C668
Quantum Mechanics
XX
=
j
i
XX
=
[cj ∗ hχj |] Â [ci |χi i]
j
cj ∗ ci hχj | Â |χi i
(8.0.9)
i
Note that  |χi i = ai |χi i and since hχj | |χi i = δi,j (since the eigenfunctions of  are
orthonormal!!) we can simplify the above equation to
D E
Â
=
XX
j
=
i
XX
j
cj ∗ ci aj hχj | |χi i
cj ∗ ci ai δi,j
(8.0.10)
i
The last term has two summations. The summation over j yields a non-zero value onlu
when j = i on account of the Kronecker delta: δi,j . Hence we have
D E
 =
X
i
ci ∗ ci ai =
X
Ni ai
(8.0.11)
i
where we have just substituted Ni = ci ∗ ci .
Now the last equation does look like an average of all the ai values. (Why? Note that
P
ci ∗ ci ≤ 1 because ψ should be normalized and that implies i ci ∗ ci = 1 from Eq.
(8.0.8).)
When you
a large number of measurements, the average of all those measurements
D perform
E
will be  . That is why it is called the “expectation value”. But a single measurement
will yield a value corresponding to one of the eigenstates
of Â!!! The average of all these
D E
measurements, just like in Eq. (8.0.11), will give  .
Consider the SG experiments and see if you can rationalize this fact. Especially consider the
experiment
You performed this experiment using the java script. Lets now look at the average value of
Sx after the second measurement. When you conduct a large number of experiments (say
10,000 runs), you get an almost equal amount of Sx+ and Sx− and the average or expectation
value turns out to be zero. (We rationalized based on the projection of the Sz+ state on Sx+
and Sx− that indeed we should get equal amounts of both.)
However, when you perform just one experiment you get either Sx+ or Sx− . Hence, the statement above, a single measurement will yield a value corresponding to one of the eigenstates
D E
of Â!!! The average of all these measurements, just like in Eq. (8.0.11), will give  .
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Course Number: C668
Quantum Mechanics
Now we are ready to find out what the Heisenberg’s Uncertainty Principle really is, in all its
glory. There are two slightly different ways to derive this and we will study both ways. The first
approach is rigorous and thats useful. The second is more elegant and leads to fun things such
as coherent states. (Incidentally Glauber got the Nobel prize for physics in 2005, for his work on
coherent states.)
Let  and B̂ be two Hermitian operators. Lets define the commutator of these operators as:
h
i
Â, B̂ = ıĈ
(8.0.12)
where Ĉ is another operator.
Homework: Prove that Ĉ is a Hermitian operator. (To prove this you will need to know that
h
ÂB̂
i†
= B̂ † † .)
Now lets define an operator D̂ that is a complex linear combination of  and B̂:
D̂ = Â + (α + ıβ) B̂
(8.0.13)
D̂ is a complex linear combinations of  and B̂. Note that α + ıβ is a complex number.
Now D̂ is an operator and hence we can say:
D̂ |Si = |Qi
(8.0.14)
That is the operator D̂ contains a recipe that converts the ket vector |Si to the ket vector |Qi
Now, the “dot” product of |Qi with itself, that is the quantity hQ |Qi ≥ 0. (Why? The dot
product of any two vectors is always greater that or equal to zero. Remind yourself that this is
indeed true by going back to Section 2.8.1 and the appendix.)
This means:
hQ |Qi = hS| D̂ † D̂ |Si ≥ 0
(8.0.15)
Note, we have used: hS| D̂ † ≡ hQ|.
If we use Eq. (8.0.13) in Eq. (8.0.15):
n
hS| Â + αB̂ + ıβ B̂
o† n
o
(8.0.16)
D E
(8.0.17)
 + αB̂ + ıβ B̂ |Si ≥ 0
Since † =  and B̂ † = B̂ since they are Hermitian, this implies
h
where Ĉ ′ = Â, B̂
fashion:
i
+
D
D
E
Â2 + α2 + β 2
D
E
D
E
B̂ 2 + α Ĉ ′ − β Ĉ ≥ 0
. Now we can rewrite the left side in the above equation in the following
Â2
E
+
D
B̂ 2
D
E

D
E 2
E2
D E2
′
1 Ĉ
1 Ĉ
D E − D E ≥0
−
4 B̂ 2
4 B̂ 2
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
D E 2
′
D E
1 Ĉ
1 Ĉ
α + D E  + B̂ 2 β − D E 
2 B̂ 2
2 B̂ 2
87
(8.0.18)
c
2014,
Srinivasan S. Iyengar (instructor)
Course Number: C668
Quantum Mechanics
Since the above expression holds for all α and β we are certainly free to choose the value of
these variables as per our convenience. In particular we choose these variables to simplify our
algebra. We choose α and β such that the bracketed terms become zero, leading to:
D
D
Â2
E
E2
D
D E2
′
1 Ĉ
1 Ĉ
− D E − D E ≥0
4 B̂ 2
4 B̂ 2
E
(8.0.19)
Multiplying by B̂ 2 , we obtain:
D
Â
ED
ED
B̂ 2 ≥
2
B̂
2
E
That is,
D
Â2
E
D
1 D ′ E2 D E2
−
Ĉ + Ĉ
≥0
4
(8.0.20)
1 D ′ E2 D E2
1 D E2
Ĉ + Ĉ
≥ Ĉ
4
4
(8.0.21)
E2
Note, we have used the fact that Ĉ ′ ≥ 0, the expectation value of a Hermitian operator is
always greater than zero.
D E2
Homework: Prove the Ĉ ′ is Hermitian operator. Also show that Ĉ ′ ≥ 0.
Now the uncertainty in the observable A is defined as:
D E
ˆ = Â − Â
∆A
(8.0.22)
Does it make sense that this is the uncertainty in A? The second term on the right side is the average
measured value or expectation value. The action of the first term on a ket yields a eigenstate of A
which could in general be different from the expectation value. (Remember the SG example from
the discussion expectation values.)
This implies the average uncertainly is given by
ˆ
∆A
2 ≡ (∆A)2 =
D E2 = Â2 − Â
D E2 = B̂ 2 − B̂
 − Â
D
E
D E2
(8.0.23)
D
E
D E2
(8.0.24)
And similarly for the operator B̂ we have:
ˆ
∆B
2 2
≡ (∆B) =
B̂ − B̂
D E
D E
If we assume the average expectation values  = B̂ = 0,
(∆A)2 (∆B)2 ≥
or
1 D E2
Ĉ
4
1 D E
Ĉ 2
Equation (8.0.26) is called the Heisenberg uncertainty principle.
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(8.0.25)
∆A∆B ≥
(8.0.26)
88
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2014,
Srinivasan S. Iyengar (instructor)
Course Number: C668
Quantum Mechanics
This equation suggests that one cannot specify, simultaneously, exact values (eigenvalues)
of a pair of non-commuting observables (e.g., position and momentum as we will see further
down below) and places quantitative restrictions on their relative variances.
Now, ∆A and ∆B uncertainties in a measurement of A and B. The equation above implies that
if the operators do not commute they cannot be simultaneously meassured with infinite certainty.
Remember we learnt earlier that commuting operators simultaneous eigenstates. When they do not
commute, their eigenstates may be different leading to the fact that they cannot be simultaneously
observed and hence the uncertainly principle above.
The essential origin of this principle is that quantum mechanics possesses the mathematical
structure of a linear vector space (viz., a Hilbert space). Note we have implicitly used nothing but
vector spaces to derive uncertainty. Why? All we assumed is operators act on kets, and yield new
kets.
1. Uncertainty of position and momentum. You may recall based on your homework that:
"
#
∂
[x̂, p̂] = x̂, −ıh̄
= ıh̄
∂x
(8.0.27)
Thus using this in Eq. (8.0.26) we obtain:
1
∆x∆p ≥ h̄
2
(8.0.28)
Thus one cannot specify, simultaneously, the position and momentum of a system. As stated
earlier this entirely due to the mathematical structure of a linear vector space that we have
been forced to introduce on account of the Stern-Gerlach experiments!! The transformation
between the position representation of the vector space and the momentum representation
is the Fourier transform as we have seen earlier, in Eq. (??), the collection of waves or
wave-packet. (exp [ıkx] is the momentum eigenstate.)
2. The Mimimum Uncertainty Wavepackets and Coherent States
There is another way to derive the Heisenberg’s uncertainty principle. We will consider that
here briefly to expose a very important concept. You may wonder Eqs. (8.0.26) and (8.0.28)
are inequalities. Are there conditions when the equality is valid? That is to put the question
a different way, what are the functions, or states, that have the minimum uncertainty product
∆x∆p = 12 h̄. We will see here that these functions are basically a “Gaussian multiplied by
a plane-wave” (or moving gaussians since the plane wave being an eigenstate of momentum
simply translates the gaussian) and are called as the minimum uncertainly wavepackets” or
“coherent states” more commonly.
Reconsider Eq. (8.0.23) in the following form:
2 ˆ
∆A
≡
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D E2 Â − Â
=
D E† D E f1 Â − Â
 −  f1
= hg1 | g1 i
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2014,
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Course Number: C668
Quantum Mechanics
=
=
D E
where |g1 i = Â − Â
Z
Z
dxg1∗ (x)g1 (x)
dx|g1 (x)|2
(8.0.29)
D E†
|f1 i and hence hg1 | = hf1 | Â − Â
In the last part of Eq. (8.0.29) we have resolved the identity using the coordinate representation. (Remember what that means? Resolution of the identity, that is 1, is inserted in the
middle in hg1 | g1 i.)
Now, we will invoke a useful mathematical tool called the Schwartz inequality which says
that for any two functions g1 and g2 :
Z
dx|g1 (x)|
2
Z
dx|g2 (x)|
2
≥
Z
2
dxg1 (x) ∗ g2 (x)
(8.0.30)
So as not to lose the flow we will accept this identity for now, but you can rationalize this
identity by considering the fact that the two terms on the left are “dot” products of g1 with
itself, and g2 with itself. The equality in Eq. (8.0.30) only holds when g2 ∝ g1 .
Since a similar equality like Eq. (8.0.29) holds
for
B̂ (that is replace  with B̂ in
D operator
E
Eq. (8.0.29) and using the function |g2 i = B̂ − B̂ |f1 i), we can say:
ˆ
∆A
2 ˆ
∆B
2 ≥
where the equality holds only when
D E† f1 Â − Â
B̂
D E 2
− B̂ f1
g2 ∝ g1
g2 = cg1
D E
D E
B̂ − B̂ |f1 i = c  −  |f1 i
(8.0.31)
(8.0.32)
If we now use x̂ and p̂ for the operators  and B̂ then
or
(x̂ − hx̂i) |f1 i = c(p̂ − hp̂i) |f1 i
(8.0.33)
!
(8.0.34)
∂
− hp̂i f1 (x)
(x − hx̂i)f1 (x) = c −ıh̄
∂x
3. Equation (8.0.34) is actually very fundamental. If choose hx̂i = hp̂i = 0 and c = 1/(ıh̄), we
get the equation
!
!
ı hp̂i
∂
f1 (x) = hx̂i +
f1 (x) = 0
(8.0.35)
x+
∂x
h̄
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Course Number: C668
Quantum Mechanics
You can see that if you pick:
(x − hx̂i)2
ı hp̂i x
f1 (x) = c1 exp −
exp
2
h̄
4 {∆x}
(
)
(
)
(8.0.36)
that would satisfy Eq. (8.0.35). (Show this for homework Comment on what ∆x must be
for this to be true. Remember we have picked hx̂i = hp̂i = 0. Eq. (8.0.36) is a gaussian
multiplied by a plane
and is called a coherent state. The operator on the left hand
wave ∂
side of Eq. (8.0.35), x + ∂x is called the annihilation operator which we will come across
when we do the Harmonic oscillator problem. It annihilates the state f1 (x) and the result is
zero.
4. Hence the coherent state is fundamental. It has the minimum uncertainty product. And hence
is the most classical-like function in quantum mechanics!! (Note in classical mechanics there
is no uncertainty and hence a function that has the minimum value is closest to classical
mechanics.) In addition the coherent state is also the eigenstate of the annihilation operator
(which we have not discussed yet).
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