What are free particles in quantum mechanics

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1. What are free particles in quantum
mechanics
The free particle is the simplest situation to apply
the Schrodinger equation.However the physics is very
interesting and there are many important consequences
of the result.
1
A particle is said to be free particle when it moves in a
regionwherethepotentialisaconstant. i.eV(x)=Constant
.In such a case the force acting on the particle isF=
dV
dx
= 0we are considering here the one dimesnional
case.
2
and a particle in box or a particle in innite
square well ?
The particle in a box and particle in innite square well
are both examples where the particles are moving in
constant potentialV(x) = 0.However the dierence is
that they are bounded by the walls of the box and and
1

2. the well. The free particle is really free so that there
are no boundary conditions.
3
sical case
In the classical situation the free particle experiences no
force and remains in a state of rest or constant motion.
4
tum mechanical situation?
To study the quantum mechanical situation we need to
solve the one dimensional time independent Schrodinger
equation.
5
tion for a free particle?
The Schrodinger equation will be written withV= 0
and will looklike as shown below

6h
2
2m
d
2

dx
2=E (1)
2

3. 6
cated!!!
Absolutely not, it is a dierential equation and can be
solved easily.For making the things simple we need to
substitute some parameters. We make the following re-
placements
k=
p
2mE
6h
(2)
so that our equation (1) becomes
d
2

dx
2=k
2
(3)
7
there any physics in these steps?
Ofcoursethesemathematicalstepshaveaphysicalmean-
ing. See carefully the equation (2). Equation (2) is
theDe-Broglie wavelengthof the particle. In order
to understand the quantum mechanical behavior of the
particle we have to understand the wave nature of the
particle.
3

4. 8
can be written in exponential form or in the
form of sines and cosines ?
The solution will be of the form
(x) = (x)e
iEt
6h(4)
where (x)is the solution and (x)are the eigen
functions.It is right that we can use the exponential
and sine cosine solutions both. However for the time
being we consider the exponential solution becausethe
free particle is unbound. The sine and cosine func-
tion are useful for the innite square well and box po-
tential where there are boundaries.
In equation (4) the quantityEdenes the energy of the
particle. The energy values are calledeigenvalues.
for an unbound solution the acceptable solutions exists
for all energy valuesE0.This physically means that
there are unrestricted values of the Energy and wave
number K.
4

5. 9
using the exponential solution ?
well let us see what the solution looks like and what in-
formation it gives regarding the particle. we write the
equation in exponential form
(x;t) =Ae
ikx
+Be
ikx
(5)
In equation (5) we have space dependence but the
time dependence is missing so we have to include the
time dependence
(x;t) =Ae
ik(x
6hkt
2m
)
+Be
ik(x+
6hkt
2m
)
(6)
Now the quantity
6hkt
2m
must be creating a confusion.Lets
see what it is
The time dependence is included ine
iEt
6hThe quantity
6hkt
2m
has to be broken to get the meaning
6h
2m

p
2mE
6h
(7)
equation (7) simplies tov. The velocity of the par-
ticle. Now our solution looks like
(x;t) =Ae
ik(xvt)
+Be
ik(x+vt)
(8)
It can be easily seen that the equation (8)
resents a traveling wave. + sign represents wave
traveling from left to right and + sign represents
wave travelling from ght to left
5

6. Figure 1: free particle traveling wave
10
The wave described above is atraveling wave. A
particular point on the wave represents a xed value of
the argument(xvt). The condition of the travelling
wave is that
propagates
argumentxvtis a constant.
xvt=constantor
x=vt+constantThe traveling wave is shown in
gure (1)
6

7. 11
like speed, wavelength etc
Remember the wave is the matter wave or De-Broglie
wave that is associated with the particle.However there
are some interesting points. First let us consider the
direction of the wave, the direction of wave is decided
by the sign of the wave vectork
k=
p
2mE
6h
(9)
k >0wavemovingright
k <0wavemovingleft
thus we can write the solution in terms of k as
(x;t) =Ae
ik(x
6hk
2
t
2m
)
(10)
Comparing with a general wave equation we can see
that that the quantity
6hk
2
2m
is analogous to frequency of
the wave . The wavelength is
=
2
k
(11) . The momentum carried by the wave is
given by the De Broglie equation.p=
6h
p
But the veloc-
ity has a discrepancy. The speed of the wave isgiven by
vquantum=
6hk
2m
(12).
Equation (12) is a general expression for nding the
speed of any travelling wave.
In classical case the speed of a moving free particle in
terms of its energy which is totally Kinetic is
7

8. 1
2
mv
2
=E-(12) or
vclassical=
q
2E
m
(13)
12
tum velocities are dierent. How to explain
this?
Yes the results are dierentvclassical= 2vquantum.I hope
thisproblemoccurredearlierinthissectionalso.Physically
this means that the quantum mechanical wave moves at
a half a speed than the particle.
13
no use? we need to nd other solution
No it is not so. The solution (8) is still a solution of the
Schrodinger equation. (x) =Ae
ikx
+Be
ikx
d
2

dx
2=k
2
(x) =
2mE
6h
2 (x)
so no doubt equation (8) is the solution
but the solution is not normalizable
The reason is that the particle isfreeit has no bound-
aries. When the solution is not normalizable then it
means that the free particlecan not have steady
solutions with xed energy.If we try to normal-
ize the wave function by conventional method
8

9. R
+1
1


(x) (x)dx=A
2
(1)(14)
Physically this means that there cannot be a
free particle whose energy is xed or quan-
tizedor
Quantization of energy occurs in bounded sys-
tems only
14
the free particle case is closed here and no
further study is required?
No, in physics when we have no solution by conventional
method then we nd new methods with new physics.
The separable solutions are still useful.The solution of
the time dependent Schrodinger equation is still a linear
combination but in a dierent way. In case of bounded
systems we have the indexnthat characterizes dier-
ent energy levels. In this case of a free particle we have
traveling waves moving from left to right and right to
left. Each such wave is characterized by the quantity
k. Soinstead of writing
(x;t) =cn (x;t)we write
(x;t) =
1
p
2
R
+1
1
(k)e
i(kx
6hk
2
2m
t)
dk(14)
9

10. 15
physical information it carries?
Theequationcontainsvitalinformationrichinphysics.The
main features of the equation (14) are
ˆThe factor
1
p
2
is for mathematical convenience and
plays the role ofcn
ˆThe variable over which the integration is carried
over the variable .
ˆThe range ofkwhich is also the range of energy
and range forms an entity calledwave packet
The general problem then becomes nding the form
of the wave function at a timetwhen the wave function
at t=0 is given . Now concentrate on the physics here .
We saw that the integration is carried over the variable
k which has dierent range of energies and momentum.
So nding the wave function at a later time involves
determinig the wave packet function(k)of equation
(14).This means we need to evaluate the following
(k) =
1
p
2
R
+1
1
(x;0)e
ikx
dx-(15)
16
Equation (15 ) comes from equation (14)
10

11. R
+1
1


(x) (x)dx=A
2
(1)(14)
by the application of a mathematical toolPlancherel's
theoremwhich is the result fromFourier's analysis.
If we have a function
f(x) =
1
p
2
R
+1
1
F(k)e
ikx
dx(16)
equation (16) physically means that we are
shifting from x(coordinated space tokspace)
F(k) is called the Fourier Transform of f(x).
f(x) is called the inverse Fourier transform of
F(k)
17
learnt till now?
Suppose we have a free particle, which is initially con-
ned in the rangea < x < a.The particle is released
at time t=0. At timet= 0the wave function is as
follows
(x;0) =Aifa < x < a
(x;0) = 0if A and a are constants then nd (x;t)
Therst thing we need to do is to normalize (x;0).
This we can do using the conventional method of nor-
malization
R
+a
a
j (x;0)j
2
dx=jAj
2
R
+a
a
dx= 2ajAj
2
=A=
1
p
2a

-(17)
so we have evaluated the constant, which is possible
11

12. only because the free particle is conned in a boundary.
Now we will apply the Plancherel's theorem and use
equation (15)
(k) =
1
p
2a
1
p
2
R
+a
a
e
ikx
dx=
1
2
p
pia
e
ikx
ik
j
+a
a=
1
p
a
sin
(ka)
k

(18)
with this value of(k)we can nd the solution (x;t)
as follows
(x;t) =
1
p
2a
R
+1
1
sin(ka)
k
e
i(kx
6hk
2
2m
t)
dk(19)
Intheequation(19)thequantityofsignicancewhich
is encountered in many topics of physics is
sin(ka)
a
. This equation is perhaps in optics under the
heading of diraction .
18
the quantity involving sine?
The quantity
sin(ka)
a
is very important in terms of
physics. Mathematically the equation isnot very
easy to solvebut let us see some limiting cases.
ˆa very small: If the size of the boundaryais
very small thensin(ka)kaand
(k) =
p
a

(20)
the equation (20) is at curve which is due to uncer-
tainty principle.ais position andkis momentum.
If spread inais large then the spread inkis small.
ˆIfais very large,the spread in position in large then
12

13. the spread in momentumkis very small. In this
case
(k) =
p
a

sin(ka)
ka
.Now from mathematics we know that a function of type
sin(x)
x
is minimum at x=0 and drops to zero atx=
. The mathematical result is consistent with the
Uncertainty principle which says forlargeathe spread
inkwill be small.
19
concept explained above?
The diagrammatic representation of the concept dis-
cussed above is given below
In the gure the rectangle shows the variation of prob-
ability density withxat timet= 0the curve shows the
probability density at timet=
ma
2
6h
20
tity k(x;0)?
The quantity as k(x;0)as usual has no physical
signicance,it is the square of (x;0)that has physi-
cal meaning.What we need is information regarding the
particle which cannot be directly obtained from k(x;0)
which is the wave packet.We need to nd information
regarding the speed of the speed of the particle from
13

14. Figure 2: Figure showing the variation of probability density in space at
dierent times
14

15. Figure showing the plot of (x;0) with x
15

16. wave packet.
21
Packet means an enclosure that contains some thing
inside it. If we have to put something inside a packet
then the size of the object should be according to size
of the packet. In case the object does not t inside, we
have tomodulate or slightly change the size of
the object. In this casea wave packet is a su-
perposition of sinosodial functions whose ampli-
tude is modulated by a factor, the sinosodials
orripplesare contained in an envelope.The sit-
uation looks like as shown in the gure
22
explanation of the wave packet?
Yes it is necessary to have a deeper look into the con-
cept of wave packet.It can be emphasized thatthe con-
cept of wave packets is the result of uncertainty
principle. It will be very interesting to see where does
the uncertainty principle come from. This requires no
separate knowledge but some points related to the wave
function. Three most important aspects of are :
ˆ can interfere with itself
ˆ is maximum at the point where the particle is
present
16

17. Figure 3: Figure showing the wave packet
17

18. Figure showing (x; t)as a superposition of waves
ˆ represent single particle or photon it never rep-
resents a large number of particle or photons
Thus if we concentrate on these points we nd that the
quantity contains the behavior of a simple particle or
photon dependsonthefactorsxandt. (x;t)isawavethatcarriesinformationregardingtheparticle:TheFourieranalysisof
(x;t)shows that it is a superposition of harmonic
waves. The two gures below will illustrate this. In
the above gure the Fourier transform of (x;t)with
average wavelength0in a region of space4x. On the
other hand the Fourier transform of (x;t)is shown
below The above gure shows the Fourier transform of
18

19. Figure showing the Fourier Transform of (x; t)with respect tok
19

20. (x;t)against wave numberkwherek=
2

. From
basic mathematical tools we have
4k4x1(21)
returning to the De-Broglie wavelength
4p=4(
h

) =
h
2
4k=h4k(22)
combining equations (21) and (22) we reach to the un-
certainty relation
4x4p6h(23)
23
and the velocity of the wave?
Let us see the gure below again
20

21. wave packet containing the simple harmonic oscil-
lations or ripplesenclosed in an envelope (k)
The velocity of the particle is the velocity of the
envelope and is calledgroup velocity
The velocity of the ripples inside the envelope
and is calledgroup velocity
24
velocity of the particles,is there any mathe-
matical proof??
Yes there is a specic formalism involving Taylor
expansion but we will use a very simple method . we
have the relationE=h(23)
The group velocity is the velocity of the center of the wave and is given by
d
d
1

=
dE
dp
=
d(
p
2
2m
)
dp
=
p
m
(24)
equation (24) is the velocity of the particle.We have calculated speed of the
particle very easily from a relation called
!
k
21