3.4 Heisenberg`s uncertainty principle

3.4 Heisenberg’s uncertainty principle
The above relationship
[ X, P] = ih̄ 6= 0
has important physical consequences. We know that the commutator of the position
and momentum operator not being equal to zero means that
• The respective eigenbasis are not the same
• That these physical quantities of a particle cannot be simulatenously measured
• measuring one, interferes with measuring the other
Now let’s look at another consequence. To this end, we need to recall a little bit of
vector analysis. Let’s assume we have an ordinary vector space R n and two arbitrary
vectors a and b and a scalar product defined on the vectorsspace such that a · b is the
result of such a scalar product which we can use to define the length of a vector by
|a|2 = a · a.
Then the so called triangle inequality is fullfilled
|a| + |b|
|a + b|
which we can understand geometrically by drawing the triangle that has a, b and a + b
as sides, as in the following drawing:
.
Now from the triangle inequality another inequalty follows. We can sqaure both
sides of the above equation and get
|a + b|2  |a|2 + |b|2 + 2|a||b|
(a + b) · (a + b)  ...
|a|2 + |b|2 + 2a · b  |a|2 + |b|2 + 2|a||b|
From which it follows that
a · b  |a||b|
Now if a · b < 0 this is trivially fullfilled. But replacing b with
fullfilled and hence
|a · b|  |a||b|
b it must also be
This is the Cauchy Schwarz inequality. What does this equation look like in a complex
Hilbert space, where we deal with ket vectors | ai for example. It turns out that the
triangle inequality also hold in Hilbert spaces where the length of a vector is defined
according to
| a |2 = h a | a i
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and the dot product according to
h a|bi = hb| ai
The dot product is in general a complex number. So given the triangle inequality we
can write
q
q
q
(h a| + hb|)(| ai + |bi)
h a| ai + hb|bi
Squaring this we find
q
q
h a| ai + hb|bi + 2 h a| ai hb|bi
| h a| ai + hb|bi + h a|bi + hb| ai |
which implies
1
| h a|bi + hb| ai |
2
Let’s keep this important inequality in mind.
| a||b|
3.4.1 The variance in measurements
Let’s go back to the general properties of physical quantities in QM. We discussed that
when a QM system is in a qm state described by a vector |yi and we have an obserable
A then the expecation value of a measurement is given by
h Ai = hy| A|yi
which, if we have a discrete sprectral decomposition of A
A=
is given by
 ln |ni hn|
n
h A i = Â l n | h y | n i |2
n
If we measure position and have
X=
ˆ
dx x | x i h x |
we get the famiiar
hXi =
where
ˆ
dx x w( x )
w ( x ) = | h y | x i |2 = y ? ( x ) y ( x )
is the probability of finding a particle at x if it is in state |yi. We can think of w( x ) as a
non-negative curve defined on x. Now h X i gives us the typical value of a position measurement. How about this question. How far is a measurement of position expected to
be away from the expectation value. In other words, what is the variance in position.
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When we have a probability distribution in statistics, that is for example a bell shaped
curve, it has a typical peak value which describes the expectation value, and a width,
which describes the variablity of the distribution. In statistics this is typically quantified
by the root mean square. For example if our probability distribution is p( x ) and the
mean is defined by
ˆ
x=
dx x p( x )
the width if the distribution is typically definend by
ˆ
2
( x x ) = dx ( x x )2 p( x )
which we can also write like
0  (x
= x2
x )2
x2
Now, how could we translate this into a quantum mechanical system. Let’s first work
with the general operator A. The expectation value is
h Ai = hy| A|yi
This is just a number, the typical measurement value of A. Let’s define the operator
DA = A
h Ai
This operator is just like A but subtracted the expectation value of A. It’s easy to show
that
hDAi = 0.
But how about the operator
DA2 = ( A
= A2
= A2
h Ai)2
A h Ai
h A i2
h A i A + h A i2
The expectation value of this operator, spits out numbers that tell us how broad the
distribution of measurement values of A are
⌦
↵ ⌦ ↵
DA2 = A2
h A i2 .
For the position operator we get
⌦
↵ ⌦
↵
X 2 = y| X 2 |y
ˆ
= dx x2 w( x )
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so
⌦
with, again,
↵
DX 2 =
Let’s use the following notation
ˆ
h X i)2 w( x )
dx ( x
w( x ) = |y( x )|2 .
µx = hX i
and
⌦
↵
sx2 = DX 2
We can do the same for the momentum, for instance compute the expectation value
ˆ
µ p = h Pi = dp p w̃( p)
where
w̃( p) = | hy| pi |2 = ỹ? ( p)ỹ( p)
is the probabilty of a particle in state |yi having a momentum p. We can also define the
mean square discplacment of of in analgoy to what we did for position
⌦
↵ ⌦
↵
sp2 = DP2 = ( P h Pi)2
Now. Let’s recall this, the state of the system is defined by the unique vector |yi and
this vector produces both, the wave functions
y( x )
and
ỹ( p)
w( x )
and
w̃( p)
and thus the probabilities
That means, in general the numbers
µ x , µ p , sx , sp
depend on one another, i.e. if we change the state |yi all for numbers change.
Now, having this in mind, let’s look at two arbitrary observables A and B and their
associated difference operators
DA = A
DB = B
h Ai
h Bi
Let’s compute the commutator of these two difference operators
[DA, DB] = DADB DBDA
= ( A h Ai) ( B h Bi) ( B h Bi) ( A h Ai)
= AB µ a B µb A + µ a µb BA + µb A + µ a B
= AB BA
= [ A, B]
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µ a µb
ha. this means that commutator of A and B are identical to the commutator of their
difference operators. This also means that for example
[DXDP] = ih̄.
because
[ X, P] = ih̄
So Now, let’s recall the Cauchy Schwarz inequality
| a||b|
1
| h a|bi + hb| ai |
2
For arbitrary vectors | ai and |bi in a Hilbert space. Let now define new vectors
| ai = A |yi
|bi = i B |yi
With this we have
⌦
↵
h a | a i = y | A2 | y
⌦
↵
h b | b i = y | B2 | y
and
h a|bi = i hy| AB|yi
and
hb| ai =
i hy| BA|yi
Now plugged into the CS inequality we get
q
1
|i (hy| AB|yi
2
1
hy|[ A, B]|yi
2
h A2 i h B2 i
hy| BA|yi)|
Now let’s apply this first to two difference operators
DA, DB
where we get
q
1
hy|[ A, B]|yi
2
hDA2 i hDB2 i
p
what was DA2 again? The expected deviation from the mean of A. Applied to position and momentum X,P we get
q
hDX 2 i hDP2 i
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h̄
2
and we can write
h̄
2
This is called Heisenbergs uncertainty principle. Remember that sx quantifies the width
of the probability distribution w( x ) = |y( x )|2 and sp that of the probability distribution
w̃( p). The uncertainty principle states that if the width of w( x ) is small then that of p
must be large and vice versa.
sx sp
3.4.1.1 Example
Let’s look at this with a specific example. Let’s assume that we have prepared the
system in a state |yi such that the probability of a particle having a momentum p is
given by
(
(2p0 ) 1
| p | < p0
w̃( p) =
0
otherwise
This means the a particle has a uniform probabiliy of having momentum p on the intervale p0 , p0 . A possible wave function in p representation
1
h p|yi = ỹ( p) = p
2p0
because this way we have
ỹ? ( p)ỹ( p) = w( p) =
1
2p0
on the same interval. The wave function in p space is in a box.
What is the expectation value of the momentum in this state?
h Pi = hy| P|yi
ˆ
= dp p w( p)
=0
because of symmetry. How about
⌦
DP
2
↵
⌦
2
↵
ˆ
= P = dp p2 w( p)
ˆ p0
1
=
dp p2 /(2p0 ) = p20
3
p0
So the width of the p distribution is proportional to p0 . Now, how about the shape of
the probability distribution of position in such a state. We have
1
h p|yi = p
2p0
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on the interval
p0 , p0 . How do we compute
y( x ) = h x |yi
We have to insert a ’one’:
y(x)=
ˆ
dp h x | pi h p|yi
ˆ
1
= dp p
eikp/h̄ y( p)
2ph̄
This is an integral that we can compute
ˆ p0
1
p
y( x ) = p
exp (ipx/h̄)
2ph̄ 2p0
p0
✓
⌘◆
1
1 ⇣ ip0 x/h̄
ip0 x/h̄
= p
e
e
2 ph̄p0 ix/h̄
✓
⇣
⌘◆
p0
1
ip0 x/h̄
ip0 x/h̄
=p
e
e
ph̄p0 2ixp0 /h̄
1
=p
1
sin( p0 x/h̄)
ph̄/p0 p0 x/h̄
So from this we get a probabilty density function
w( x ) = |y( x )|2 =
We can check that
ˆ
1 sin2 ( p0 x/h̄)
ph̄/p0 ( p0 x/h̄)2
dxw( x ) = 1
We can immediately see that the expectation value of position in this state is zero because w( x ) is symmetric. So
ˆ x0
⌦ 2↵ ⌦ 2↵
2
sin2 ( p0 x/h̄)
DX = X
dxx2
ph̄/p0 0
( p0 x/h̄)2
where
x0 = ph̄/p0
So
⌦
X
2
2
ph̄/p0
↵
=
✓
1
p0 /h̄
◆2 ˆ
2 1 1 ph̄
=
p p0 /h̄ 2 p0
This means
sx sp
✓
h̄ p0
p
p0 3
as required.
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0
h̄
p0
x0
dx sin2 ( p0 x/h̄)
◆2
h̄/2
3.5 Time evolution and the Schrödinger equation
So far we discussed situation that did not involve time. For instance the states of qm
systems have always been considered as stationary situation described by the state vector
|yi
that describe a system as it goes in to a measurement apparatus and exits in an eigenstate of that apparatus. We also discussed what happens if we sequentially let the
system go through a sequence of measurement apparati. What we implicitly assumed
here is that essentiall nothing happens to a state in between. It turns out however that
the state vector does evolve in time between measurements and interestingly is does so
deterministically.
So first we have to realize that the state vector chances over time. We may write this
according to
|yi = |y, ti
When we use a position representation we write this as
h x |y, ti = y( x, t)
This is the wave function of a particle at position x and the probability of finding a
particle at x is also a function of time
w( x, t) = |y( x, t)|2 .
Now, because the state vector changes over time, a question of fundamental importance
is according to what rule does it change, e.g. what is the right side of this equation
∂t |y, ti =?
or
∂t y( x, t) =?
Here’s how this works. Remember that when we discussed momentum we saw that
the momentum operator applied to a state vector was equivalent to the first derivative
in position representation:
P $ ( ih̄)∂ x
Explicity, let’s recall if
Then
|q i = P|yi
q (x) =
ih̄∂ x y( x )
So there’s a deep connection between momentum and change with respect to position.
A similar connection exists between energy and time which we can write in the following way
ih̄∂t $ H
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where His the Hamilton Operator, the operator that corresponds to energy. In otherworks if we build a machine that measures the energy of a qm system, it represents the
operator H the energy operator. If this is true, and we will discuss this firther below,
we may expect that if we have a time dependent state |y, ti we can write
ih̄∂t |y, ti = H |y, ti
and this is indeed the case. This equation is a variant of Schrödinger’s equation. It
describes how a time dependent state changes over time. All you need to do is act on
the state with the Hamilton operator, the energy operator. Let’s again look at this in
position repreentation:
ih̄ h x |y, ti = h x | H |y, ti
we can insert a one:
ih̄y( x, t) =
ˆ
dyH ( x, y)y(y, t)
where
H ( x, y) = h x | H |yi
is the position represntation of the Hamilton operator. Let’s look at an example. We
need to figure out how to deal with energy. Now we can use something that is called
the correspondance principle. In classical mechancis the energy of a particle is typically
a scalar function of the particles momentum and position, for instance a particle that
moves in a 1d potential U ( x ) that is a function of position is given by
H=
p2
+ U ( x ).
2m
When we have such a Hamiltonian the classical equations of motion are given by
ẋ = ∂H/∂p = p/m
ṗ =
∂H/∂x =
dU ( x )/dx
and because the derivative of the Potential energy U is the force we have
ṗ = F ( x ) = m ẍ
Now the correspondence principle states that the quantum mechanical Hamilton Operator is given by the classical one in which classical quantities are replaced by their
quantum mechanical, corresponding, operator
p!P
x!X
So the qm Hamiltonian is given by
H=
P2
+ U (X)
2m
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Now in order to rewrite the equation
ih̄y( x, t) =
ˆ
dyH ( x, y)y(y, t)
we need the position representation of the Hamiltonian.
H ( x, y) = h x | H |yi =
↵
1 ⌦
x | P2 |y + h x |U ( X )|yi
2m
Let’s first consider the last term. We know that
h x | X |yi = xd( x
y)
And we can show that any power of X e.g. X n gives
h x | X |n yi = x n d( x
y)
Now we could replace U ( X ) by a series of powers in X and we will find that
h x |U ( X )|yi = U ( x )d( x
y)
In face the operation of U ( X ) on any state |yi in position representation just means the
multiplication with the function U ( x ), if we have
U ( X ) |yi = |q i
then
h x |U ( X )|yi = q ( x )
U ( x )y( x ) = q ( x )
Now how about this part
⌦
Let’s do this step by step
↵
x | P2 | y ?
h x | PP|yi =
ˆ
dz h x | P|zi hz| P|yi
=
ˆ
dzd( x
=
h̄2 d( x
z)( ih̄∂z )d(z
y)( ih̄∂y )
y)∂2y
In fact we can always make the translation
Pn $ ( ih̄)n ∂nx
in position representation. So, now we have
H ( x, y) = d( x
h
y) U (y)
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h̄2 ∂2y
i
and we get from
ih̄y( x, t) =
ˆ
dyH ( x, y)y(y, t)
the most important equation in quantum mechanics, the Schrödinger Equation
h̄2 2
∂ y( x, t) + U ( x )y( x, t)
2m x
of a massive, non-relativistic particle in a potential. The solution of this equation gives
the time dependent wave function from which we can compute the time dependent
probability of finding a particle at some position x. In order to solve this, we only need
to specify the state of the system at time t = 0, which is y( x, 0).
Let’s consider a free particle of mass m first. In this case U ( x ) = 0 and the Schrödinger
equation reads
h̄2 2
ih̄∂t y( x, t) =
∂ y( x, t)
2m x
Remember that this is the position representation of
ih̄∂t y( x, t) =
1 2
P |yi
2m
that we get when we multiple from the left with h x |. It turns out that solving the above
equation is easier in momentum representation when we multiply with h p|then we get
ih̄∂t |yi =
writing
we get
ih̄∂t h p|y, ti =
1 2
p h p|y, ti
2m
h p|y, ti = ỹ( p, t)
ih̄∂t ỹ( p, t) =
This has the solution
p2
ỹ( p, t)
2m
2
y( p, t) = y( p, 0)e (ip /2h̄m)t
Now
y( p, 0) = h p|y, 0i
is the probability amplitude of the particle having a momentum p at time t = 0. and
y( p, t) = h p|y, ti
is the amplitude of the particle having a momentum at time t. How can we compute
the wave function y( x, t) from this? Well we have
ˆ
y( x, t) = h x |y, ti = dp h x | pi h p|y, ti
ˆ
2
1
=p
dp exp (ipx/h̄) y( p, 0)e (ip /2h̄m)t
2ph̄
 ✓
◆
ˆ
1
i
p2
=p
dpy( p, 0) exp
px
t
h̄
2m
2ph̄
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So, we see that this is a superposition of travelling waves with
h̄k = p
h̄w = E = p2 /2m.
Let’s calculate this for a given intitial function
y( p, 0) = A exp

p0 )2 s 2
(p
h̄2
Plugging this into the above equation we get
y( x, t) = p
A
2ph̄
ˆ
 ✓
i
dp exp
px
h̄
p2
t
2m
(p
p0 )2 s 2
h̄2
◆
This integral is a bit cmplicated to evaluate but it will give
y( x, t) = C exp
with˘
b=
s2 p20
2
+ ix/2h̄
a=
s2
2

b2
a
d
+i
t
2mh̄
d = s2 p20 /h̄2
h̄
h̄
This is very complicated to illustrate. But what we are really interested in is the probabiliy
✓
◆
1
( x vt)2
w( x, t) = |y( x, t)|2 = p
exp
2s2 (1 + D2 )
s 2p (1 + D2 )
with
v = p0 /m
D = th̄/2ms2
This describes a wave paket that is moving to the right at a speed v that is spreading
out, getting wider an wider.
3.5.1 Time independent Schördinger Equation
Now let’s look at the Schrödinger Equation with a potential again, either in the general
form
ih̄∂t |y, ti = H |y, ti
or the position representation
ih̄∂t y( x, t) =
h̄2 2
∂ y( x, t) + U ( x )y( x, t)
2m x
We can make the following ansatz
|y, ti = e
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iEt/h̄
|fi
plugged into the time-dependent SE we obtain
iE
|fi = H |fi
h̄
ih̄
so
E |fi = H |fi
This means that the constant E is an eigenvalue of the Hamilton Operator H. Once we
compute this and the associated eigenvector |fi we have the general time-dependent
solution
|y, ti = e iEt/h̄ |fi
So now we have to first find the set of eigenvalues and eigenfunction for a given Hamilton Operators.
3.5.2 Let’s look at the Harmonic Oscillator
The Harmonic oscillator describes a massive particle that is subject to s linear force
F(x) =
kx
where w is a positive constant. Then Newton tells us that
k
x
m
ẍ =
This has solutions that are harmonic, e.g.
p
x (t) = x (0) cos(t k/m)
So oscillations with a frequency
w=
p
k/m
Now the classical Hamiltonian for this looks like
H=
p2
x2
+k
2m
2
and we can also derive the equations of motion using
ẋ = ∂H/∂p = p/m
and
ṗ =
∂H/∂x =
kx
which are equivalent to the above. Using the correspondance principle we have a
Hamiltonian
P2
1
+ kX 2
2m 2
P2
mw 2 2
=
+
X
2m
2
H=
76
Now, as described above, we first want to find the eigenfunctions of H according do
H |fi = E |fi
In position representation this equation reads
"
#
h̄2 2 mw 2 2
∂ +
x f( x ) = Ef( x )
2m x
2
or

h̄2 00
f (x) = E
2m
mw 2 x2
f( x )
2
This is a differential equation for which we need to find a solution. We could use the
typical differential equation bla bla to do this. However, we will go along a different
route which will be based on operators and algebra. So, let’s go back to
H |fi = E |fi
that is

P2
mw 2 2
+
X |fi = E |fi
2m
2
Now consider the following operators with
r
x0 =
and its conjugate
1
a= p
2
✓
1
a = p
2
✓
†
h̄
wm
1
i
X + x0 P
x0
h̄
1
X
x0
◆
i
x0 P
h̄
◆
This operators are not Hermitian. We can express position and momentum in terms of
a and a†
⇣
⌘
1
X = p x0 a + a †
2
⌘
1 ⇣
P = ih̄ p
a a†
2x0
Now, it’s easy to show that
[ a, a† ] = 1
What do these operators look like in position representation, let
a |yi = |q i
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so if we multiply with a h x | from the left we get
✓
◆
1
x
q (x) = p
+ x0 ∂ x y ( x )
2 x0
and similarly for a† where the derivative term has a minus sign in front of it. Now if
we plug in the above equations in the definition of H we get
1
h̄w ( a† a + aa† )
2
H=
but
1 = aa†
a† a
so
1
h̄w (2a† a + 1)
2
= h̄w ( a† a + 1/2)
H=
Let’s define the operator
n = a† a.
This operator is Hermitian. So we have
H = h̄w (n + 1/2).
Now, if we find the eigenvalues and eigenfunctions of n we will have the eigenfunctions
and eigenvalues of H. So all we need to solve is this eigenvalue problem
n |ni = n |ni .
Now let’s try to find out something about the eigenvalues. If we multiply by hn|we get
D
hn|n|ni = n
E
n| a† a|n = n
| a | n i |2 = n
and thus
0
n
So all possible eigenvalues must be non-negative. Let’s see if we assume that n = 0.
Then we have
n |ni = 0
That must also imply that the norm of a |ni is zero which means that
a |ni = 0
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which reads in position representation
✓
◆
x
+ x0 ∂ x n ( x ) = 0
x0
For which we find the solution
n0 ( x ) =
✓
1
p
px0
◆1/2
exp
1
2
✓
x
x0
◆2 !
.
This means that this is an eigenfuntion of H with eigenvalue
E = h̄w/2 > 0
This is interesting. This means that there is no state, and hence no wave function that
has a lower energy. The minimal energy of a particle in a quadratic potential is positive
and not zero. Quite fascinating. In a way, the qm particle is never at rest.
Now, how about the other eigenfunctions and eigenvalues of H? Well, if |ni is an
eigenstate of n with eigenvalue n then the state
| n + 1i = p
1
a† |ni
n+1
is also a normalized eigenstate with eigenvalue n + 1, i.e.
na† |ni = (n + 1) a† |ni
To show this we need to check first that
[n, a† ] = a† = na†
a† n
So
⇣
⌘
na† |ni = a† n + a† |ni
= ( n + 1) a † | n i
Is it normalized
E
1 D
n| aa† |n
n+1
E
1 D
=
n |1 + a † a | n
n+1
=1
h n + 1| n + 1i =
So that means, we can start with the ground state |0i
• Time independent SE
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