3.4 Heisenberg’s uncertainty principle

The above relationship[X, P] = ih 6= 0

has important physical consequences. We know that the commutator of the positionand 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 ofvector analysis. Let’s assume we have an ordinary vector space Rn and two arbitraryvectors a and b and a scalar product defined on the vectorsspace such that a · b is theresult 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 + bas 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 �b it must also befullfilled and hence

|a · b| |a||b|This is the Cauchy Schwarz inequality. What does this equation look like in a complexHilbert space, where we deal with ket vectors |ai for example. It turns out that thetriangle inequality also hold in Hilbert spaces where the length of a vector is definedaccording to

|a|2 = ha|ai

64

and the dot product according to

ha|bi = hb|aiThe dot product is in general a complex number. So given the triangle inequality wecan write q

ha|ai+qhb|bi �

q(ha|+ hb|)(|ai+ |bi)

Squaring this we find

ha|ai+ hb|bi+ 2qha|ai

qhb|bi � | ha|ai+ hb|bi+ ha|bi+ hb|ai |

which implies

|a||b| � 12| ha|bi+ hb|ai |

Let’s keep this important inequality in mind.

3.4.1 The variance in measurements

Let’s go back to the general properties of physical quantities in QM. We discussed thatwhen a QM system is in a qm state described by a vector |yi and we have an obserableA then the expecation value of a measurement is given by

hAi = hy|A|yiwhich, if we have a discrete sprectral decomposition of A

A = Ân

ln |ni hn|

is given byhAi = Â

nln| hy|ni |2

If we measure position and have

X =ˆ

dx x |xi hx|

we get the famiiar

hXi =ˆ

dx x w(x)

wherew(x) = | hy|xi |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 anon-negative curve defined on x. Now hXi gives us the typical value of a position mea-surement. How about this question. How far is a measurement of position expected tobe away from the expectation value. In other words, what is the variance in position.

65

When we have a probability distribution in statistics, that is for example a bell shapedcurve, 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 quantifiedby the root mean square. For example if our probability distribution is p(x) and themean is defined by

x =ˆ

dx x p(x)

the width if the distribution is typically definend by

(x � x)2 =ˆ

dx (x � x)2 p(x)

which we can also write like

0 (x � x)2

= x2 � x2

Now, how could we translate this into a quantum mechanical system. Let’s first workwith the general operator A. The expectation value is

hAi = hy|A|yiThis is just a number, the typical measurement value of A. Let’s define the operator

DA = A � hAiThis operator is just like A but subtracted the expectation value of A. It’s easy to showthat

hDAi = 0.

But how about the operator

DA2 = (A � hAi)2

= A2 � A hAi � hAi A + hAi2

= A2 � hAi2

The expectation value of this operator, spits out numbers that tell us how broad thedistribution of measurement values of A are

⌦DA2↵ =

⌦A2↵� hAi2 .

For the position operator we get⌦

X2↵ =⌦y|X2|y↵

=ˆ

dx x2w(x)

66

so ⌦DX2↵ =

ˆdx(x � hXi)2w(x)

with, again,w(x) = |y(x)|2.

Let’s use the following notationµx = hXi

ands2

x =⌦DX2↵

We can do the same for the momentum, for instance compute the expectation value

µp = hPi =ˆ

dp p w(p)

wherew(p) = | hy|pi |2 = y?(p)y(p)

is the probabilty of a particle in state |yi having a momentum p. We can also define themean square discplacment of of in analgoy to what we did for position

s2p =

⌦DP2↵ =

⌦(P � hPi)2↵

Now. Let’s recall this, the state of the system is defined by the unique vector |yi andthis vector produces both, the wave functions

y(x) and y(p)

and thus the probabilitiesw(x) and w(p)

That means, in general the numbers

µx, µp, 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 � hAiDB = B � hBi

Let’s compute the commutator of these two difference operators

[DA, DB] = DADB � DBDA= (A � hAi) (B � hBi)� (B � hBi) (A � hAi)= AB � µaB � µb A + µaµb � BA + µb A + µaB � µaµb

= AB � BA= [A, B]

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ha. this means that commutator of A and B are identical to the commutator of theirdifference operators. This also means that for example

[DXDP] = ih.

because[X, P] = ih

So Now, let’s recall the Cauchy Schwarz inequality

|a||b| � 12| ha|bi+ hb|ai |

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

ha|ai = ⌦y|A2|y↵

hb|bi = ⌦y|B2|y↵

andha|bi = i hy|AB|yi

andhb|ai = �i hy|BA|yi

Now plugged into the CS inequality we getqhA2i hB2i � 1

2|i(hy|AB|yi � hy|BA|yi)|

� 12hy|[A, B]|yi

Now let’s apply this first to two difference operators

DA, DB

where we get qhDA2i hDB2i � 1

2hy|[A, B]|yi

what wasp

DA2 again? The expected deviation from the mean of A. Applied to posi-tion and momentum X,P we get

qhDX2i hDP2i � h

2

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and we can writesxsp � h

2This is called Heisenbergs uncertainty principle. Remember that sx quantifies the widthof the probability distribution w(x) = |y(x)|2 and sp that of the probability distributionw(p). The uncertainty principle states that if the width of w(x) is small then that of pmust be large and vice versa.

3.4.1.1 Example

Let’s look at this with a specific example. Let’s assume that we have prepared thesystem in a state |yi such that the probability of a particle having a momentum p isgiven by

w(p) =

((2p0)�1 |p| < p0

0 otherwise

This means the a particle has a uniform probabiliy of having momentum p on the in-tervale �p0, p0. A possible wave function in p representation

hp|yi = y(p) =1p2p0

because this way we have

y?(p)y(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?

hPi = hy|P|yi=ˆ

dp p w(p)

= 0

because of symmetry. How about

⌦DP2↵ =

⌦P2↵ =

ˆdp p2w(p)

=ˆ p0

�p0

dp p2/(2p0) =13

p20

So the width of the p distribution is proportional to p0. Now, how about the shape ofthe probability distribution of position in such a state. We have

hp|yi = 1p2p0

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on the interval �p0, p0. How do we compute

y(x) = hx|yiWe have to insert a ’one’:

y(x)=ˆ

dp hx|pi hp|yi

=ˆ

dp1p2ph

eikp/hy(p)

This is an integral that we can compute

y(x) =1p2ph

1p2p0

ˆ p0

�p0

exp (ipx/h)

=1

2p

php0

✓1

ix/h

⇣eip0x/h � e�ip0x/h

⌘◆

=p0pphp0

✓1

2ixp0/h

⇣eip0x/h � e�ip0x/h

⌘◆

=1p

ph/p0

sin(p0x/h)p0x/h

So from this we get a probabilty density function

w(x) = |y(x)|2 =1

ph/p0

sin2(p0x/h)(p0x/h)2

We can check that ˆdxw(x) = 1

We can immediately see that the expectation value of position in this state is zero be-cause w(x) is symmetric. So

⌦DX2↵ =

⌦X2↵ � 2

ph/p0

ˆ x0

0dxx2 sin2(p0x/h)

(p0x/h)2

wherex0 = ph/p0

So⌦

X2↵ � 2ph/p0

✓1

p0/h

◆2 ˆ x0

0dx sin2(p0x/h)

=2p

1p0/h

12

php0

=

✓hp0

◆2

This meanssxsp � h

p0

p0p3� h/2

as required.

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3.5 Time evolution and the Schrödinger equation

So far we discussed situation that did not involve time. For instance the states of qmsystems have always been considered as stationary situation described by the state vec-tor

|yithat describe a system as it goes in to a measurement apparatus and exits in an eigen-state of that apparatus. We also discussed what happens if we sequentially let thesystem go through a sequence of measurement apparati. What we implicitly assumedhere is that essentiall nothing happens to a state in between. It turns out however thatthe state vector does evolve in time between measurements and interestingly is does sodeterministically.

So first we have to realize that the state vector chances over time. We may write thisaccording to

|yi = |y, tiWhen we use a position representation we write this as

hx|y, ti = y(x, t)

This is the wave function of a particle at position x and the probability of finding aparticle 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 importanceis according to what rule does it change, e.g. what is the right side of this equation

∂t |y, ti =?

or∂ty(x, t) =?

Here’s how this works. Remember that when we discussed momentum we saw thatthe momentum operator applied to a state vector was equivalent to the first derivativein position representation:

P $ (�ih)∂x

Explicity, let’s recall if|qi = P|yi

Thenq(x) = �ih∂xy(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 follow-ing way

ih∂t $ H

71

where His the Hamilton Operator, the operator that corresponds to energy. In other-works if we build a machine that measures the energy of a qm system, it represents theoperator 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

ih∂t |y, ti = H |y, tiand this is indeed the case. This equation is a variant of Schrödinger’s equation. Itdescribes how a time dependent state changes over time. All you need to do is act onthe state with the Hamilton operator, the energy operator. Let’s again look at this inposition repreentation:

ih hx|y, ti = hx|H|y, tiwe can insert a one:

ihy(x, t) =ˆ

dyH(x, y)y(y, t)

whereH(x, y) = hx|H|yi

is the position represntation of the Hamilton operator. Let’s look at an example. Weneed to figure out how to deal with energy. Now we can use something that is calledthe correspondance principle. In classical mechancis the energy of a particle is typicallya scalar function of the particles momentum and position, for instance a particle thatmoves in a 1d potential U(x) that is a function of position is given by

H =p2

2m+ U(x).

When we have such a Hamiltonian the classical equations of motion are given by

x = ∂H/∂p = p/mp = �∂H/∂x = �dU(x)/dx

and because the derivative of the Potential energy U is the force we have

p = F(x) = mx

Now the correspondence principle states that the quantum mechanical Hamilton Op-erator is given by the classical one in which classical quantities are replaced by theirquantum mechanical, corresponding, operator

p ! P x ! X

So the qm Hamiltonian is given by

H =P2

2m+ U(X)

72

Now in order to rewrite the equation

ihy(x, t) =ˆ

dyH(x, y)y(y, t)

we need the position representation of the Hamiltonian.

H(x, y) = hx|H|yi = 12m

⌦x|P2|y↵+ hx|U(X)|yi

Let’s first consider the last term. We know that

hx|X|yi = xd(x � y)

And we can show that any power of X e.g. Xn gives

hx|X|nyi = xnd(x � y)

Now we could replace U(X) by a series of powers in X and we will find that

hx|U(X)|yi = U(x)d(x � y)

In face the operation of U(X) on any state |yi in position representation just means themultiplication with the function U(x), if we have

U(X) |yi = |qithen

hx|U(X)|yi = q(x)U(x)y(x) = q(x)

Now how about this part ⌦x|P2|y↵?

Let’s do this step by step

hx|PP|yi =ˆ

dz hx|P|zi hz|P|yi

=ˆ

dzd(x � z)(�ih∂z)d(z � y)(�ih∂y)

= �h2d(x � y)∂2y

In fact we can always make the translation

Pn $ (�ih)n∂nx

in position representation. So, now we have

H(x, y) = d(x � y)hU(y)� h2∂2

y

i

73

and we get from

ihy(x, t) =ˆ

dyH(x, y)y(y, t)

the most important equation in quantum mechanics, the Schrödinger Equation

ih∂ty(x, t) = � h2

2m∂2

xy(x, t) + U(x)y(x, t)

of a massive, non-relativistic particle in a potential. The solution of this equation givesthe time dependent wave function from which we can compute the time dependentprobability of finding a particle at some position x. In order to solve this, we only needto 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 Schrödingerequation reads

ih∂ty(x, t) = � h2

2m∂2

xy(x, t)

Remember that this is the position representation of

ih∂t |yi = 12m

P2 |yithat we get when we multiple from the left with hx|. It turns out that solving the aboveequation is easier in momentum representation when we multiply with hp|then we get

ih∂t hp|y, ti = 12m

p2 hp|y, tiwriting

hp|y, ti = y(p, t)

we get

ih∂ty(p, t) =p2

2my(p, t)

This has the solutiony(p, t) = y(p, 0)e�(ip2/2hm)t

Nowy(p, 0) = hp|y, 0i

is the probability amplitude of the particle having a momentum p at time t = 0. and

y(p, t) = hp|y, tiis the amplitude of the particle having a momentum at time t. How can we computethe wave function y(x, t) from this? Well we have

y(x, t) = hx|y, ti =ˆ

dp hx|pi hp|y, ti

=1p2ph

ˆdp exp (ipx/h)y(p, 0)e�(ip2/2hm)t

=1p2ph

ˆdpy(p, 0) exp

ih

✓px � p2

2mt◆�

74

So, we see that this is a superposition of travelling waves with

hk = p hw = E = p2/2m.

Let’s calculate this for a given intitial function

y(p, 0) = A exp� (p � p0)2s2

h2

�

Plugging this into the above equation we get

y(x, t) =Ap2ph

ˆdp exp

ih

✓px � p2

2mt � (p � p0)2s2

h2

◆�

This integral is a bit cmplicated to evaluate but it will give

y(x, t) = C exp

b2

a� d

�

with˘

b =s2 p2

0

h2 + ix/2h a =s2

h2 + it

2mhd = s2 p2

0/h2

This is very complicated to illustrate. But what we are really interested in is the proba-biliy

w(x, t) = |y(x, t)|2 =1

sp

2p(1 + D2)exp

✓� (x � vt)2

2s2(1 + D2)

◆

withv = p0/m D = th/2ms2

This describes a wave paket that is moving to the right at a speed v that is spreadingout, 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 generalform

ih∂t |y, ti = H |y, tior the position representation

ih∂ty(x, t) = � h2

2m∂2

xy(x, t) + U(x)y(x, t)

We can make the following ansatz

|y, ti = e�iEt/h |fi

75

plugged into the time-dependent SE we obtain

ih�iE

h|fi = H |fi

soE |fi = H |fi

This means that the constant E is an eigenvalue of the Hamilton Operator H. Once wecompute this and the associated eigenvector |fi we have the general time-dependentsolution

|y, ti = e�iEt/h |fiSo now we have to first find the set of eigenvalues and eigenfunction for a given Hamil-ton 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

x = � km

x

This has solutions that are harmonic, e.g.

x(t) = x(0) cos(tp

k/m)

So oscillations with a frequencyw =

pk/m

Now the classical Hamiltonian for this looks like

H =p2

2m+ k

x2

2and we can also derive the equations of motion using

x = ∂H/∂p = p/m

andp = �∂H/∂x = �kx

which are equivalent to the above. Using the correspondance principle we have aHamiltonian

H =P2

2m+

12

kX2

=P2

2m+

mw2

2X2

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Now, as described above, we first want to find the eigenfunctions of H according do

H |fi = E |fiIn position representation this equation reads

"� h2

2m∂2

x +mw2

2x2

#f(x) = Ef(x)

orh2

2mf00(x) =

E � mw2x2

2

�f(x)

This is a differential equation for which we need to find a solution. We could use thetypical differential equation bla bla to do this. However, we will go along a differentroute which will be based on operators and algebra. So, let’s go back to

H |fi = E |fithat is

P2

2m+

mw2

2X2

�|fi = E |fi

Now consider the following operators with

x0 =

rh

wm

a =1p2

✓1x0

X +ih

x0P◆

and its conjugate

a† =1p2

✓1x0

X � ih

x0P◆

This operators are not Hermitian. We can express position and momentum in terms ofa and a†

X =1p2

x0

⇣a + a†

⌘

P = �ih1p2x0

⇣a � a†

⌘

Now, it’s easy to show that[a, a†] = 1

What do these operators look like in position representation, let

a |yi = |qi

77

so if we multiply with a hx| from the left we get

q(x) =1p2

✓xx0

+ x0∂x

◆y(x)

and similarly for a† where the derivative term has a minus sign in front of it. Now ifwe plug in the above equations in the definition of H we get

H =12

hw(a†a + aa†)

but1 = aa† � a†a

so

H =12

hw(2a†a + 1)

= hw(a†a + 1/2)

Let’s define the operatorn = a†a.

This operator is Hermitian. So we have

H = hw(n + 1/2).

Now, if we find the eigenvalues and eigenfunctions of n we will have the eigenfunctionsand 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

hn|n|ni = nD

n|a†a|nE= n

|a |ni |2 = n

and thusn � 0

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

78

which reads in position representation✓

xx0

+ x0∂x

◆n(x) = 0

For which we find the solution

n0(x) =✓

1ppx0

◆1/2exp

�1

2

✓xx0

◆2!

.

This means that this is an eigenfuntion of H with eigenvalue

E = hw/2 > 0

This is interesting. This means that there is no state, and hence no wave function thathas a lower energy. The minimal energy of a particle in a quadratic potential is positiveand 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 aneigenstate of n with eigenvalue n then the state

|n + 1i = 1pn + 1

a† |ni

is also a normalized eigenstate with eigenvalue n + 1, i.e.

na† |ni = (n + 1)a† |niTo show this we need to check first that

[n, a†] = a† = na† � a†n

So

na† |ni =⇣

a†n + a†⌘|ni

= (n + 1)a† |niIs it normalized

hn + 1|n + 1i = 1n + 1

Dn|aa†|n

E

=1

n + 1

Dn|1 + a†a|n

E

= 1

So that means, we can start with the ground state |0i• Time independent SE

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