detailed notes and solved problems on a free particle in quantum mechanics. For details visit https://www.learnandenjoy-physics.com/

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 What is a free particle?
A particle is said to be free particle when it moves in a
region where the potential is a constant. i.eV(x)=Constant
.In such a case the force acting on the particle is F =
dV
dx
= 0 we are considering here the one dimesnional
case.
2 What is the difference between a free particle
and a particle in box or a particle in infinite
square well ?
The particle in a box and particle in infinite square well
are both examples where the particles are moving in
constant potential V (x) = 0.However the difference 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 How does a free particle behave in the clas-
sical case
In the classical situation the free particle experiences no
force and remains in a state of rest or constant motion.
4 How can we study the behavior in the quan-
tum mechanical situation?
To study the quantum mechanical situation we need to
solve the one dimensional time independent Schrodinger
equation.
5 What will be the form of Schrodinger equa-
tion for a free particle?
The Schrodinger equation will be written with V = 0
and will looklike as shown below
− 6h2
2m
d2ψ
dx2 = Eψ —(1)
2

3. 6 How to solve this equation? it looks compli-
cated!!!
Absolutely not, it is a differential equation and can be
solved easily.For making the things simple we need to
substitute some parameters. We make the following re-
placements
k =
√
2mE
6h —(2)
so that our equation (1) becomes
d2ψ
dx2 = −k2
ψ—(3)
7 Are these simply mathematical steps or is
there any physics in these steps?
Of course these mathematical steps have a physical mean-
ing. See carefully the equation (2). Equation (2) is
the De-Broglie wavelength of 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 The solution of the differential equation (1)
can be written in exponential form or in the
form of sines and cosines ?
The solution will be of the form
Ψ(x) = ψ(x)e
−iEt
6h —(4)
where Ψ(x) is the solution and ψ(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 because the
free particle is unbound. The sine and cosine func-
tion are useful for the infinite square well and box po-
tential where there are boundaries.
In equation (4) the quantity E defines the energy of the
particle. The energy values are called eigenvalues.
for an unbound solution the acceptable solutions exists
for all energy values E ≥ 0.This physically means that
there are unrestricted values of the Energy and wave
number K.
4

5. 9 What physical information can be obtained
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
Ψ(x, t) = Aeikx
+ Be−ikx
—(5)
In equation (5) we have space dependence but the
time dependence is missing so we have to include the
time dependence
Ψ(x, t) = Aeik(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 in e
−iEt
6h The quantity
6hkt
2m has to be broken to get the meaning
6h
2m ×
√
2mE
6h –(7)
equation (7) simplifies to v . The velocity of the par-
ticle. Now our solution looks like
Ψ(x, t) = Aeik(x−vt)
+ Beik(x+vt)
—(8)
It can be easily seen that the equation (8) rep-
resents a traveling wave. + sign represents wave
traveling from left to right and + sign represents
wave travelling from fight to left
5

6. Figure 1: free particle traveling wave
10 How can I visualize the wave?
The wave described above is a traveling wave. A
particular point on the wave represents a fixed value of
the argument (x − vt). The condition of the travelling
wave is that the profile of the wave remains same as it
propagates This can be mathematically shown as the
argument x − vt is a constant.
x ± vt = constant or
x = ∓vt + constant The traveling wave is shown in
figure (1)
6

7. 11 What are the other properties of the wave
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 vector k
k =
√
2mE
6h
—(9)
k > 0wavemovingright
k < 0wavemovingleft
thus we can write the solution in terms of k as
Ψ(x, t) = Aeik(x−6hk2t
2m )
—(10)
Comparing with a general wave equation we can see
that that the quantity 6hk2
2m is analogous to frequency of
the wave . The wavelength is
λ = 2π
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 finding 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
2mv2
= E—-(12) or
vclassical =
q
2E
m –(13)
12 The result is bizarre the classical and quan-
tum velocities are different. How to explain
this?
Yes the results are different vclassical = 2vquantum.I hope
this problem occurred earlier in this section also.Physically
this means that the quantum mechanical wave moves at
a half a speed than the particle.
13 This means that the solution obtained is of
no use? we need to find other solution
No it is not so. The solution (8) is still a solution of the
Schrodinger equation. ψ(x) = Aeikx
+ Be−ikx
d2ψ
dx2 = −k2
ψ(x) = 2mE
6h2 ψ(x)
so no doubt equation (8) is the solution
but the solution is not normalizable
The reason is that the particle is free it has no bound-
aries. When the solution is not normalizable then it
means that the free particle can not have steady
solutions with fixed energy. If we try to normal-
ize the wave function by conventional method
8

9. R +∞
−∞ Ψ∗
(x)Ψ(x)dx = A2
(∞)—(14)
Physically this means that there cannot be a
free particle whose energy is fixed or quan-
tized or
Quantization of energy occurs in bounded sys-
tems only
14 So if the solution is not normalizable, then
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 find 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 different way. In case of bounded
systems we have the index n that characterizes differ-
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
Ψ(x, t) = cnψ(x, t) we write
ψ(x, t) = 1
√
2π
R +∞
−∞ φ(k)ei(kx−6hk2
2m t)
dk—(14)
9

10. 15 What is meaning of this equation and what
physical information it carries?
The equation contains vital information rich in physics.The
main features of the equation (14) are
• The factor 1
√
2π
is for mathematical convenience and
plays the role of cn
• The variable over which the integration is carried
over the variable .
• The range of k which is also the range of energy
and range forms an entity called wave packet
The general problem then becomes finding the form
of the wave function at a time t when 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 different range of energies and momentum.
So finding the wave function at a later time involves
determinig the wave packet function φ(k) of equation
(14).This means we need to evaluate the following
φ(k) = 1
√
2π
R +∞
−∞ ψ(x, 0)e−ikx
dx—-(15)
16 Where does the equation (15) come from?
Equation (15 ) comes from equation (14)
10

11. R +∞
−∞ Ψ∗
(x)Ψ(x)dx = A2
(∞)—(14)
by the application of a mathematical tool Plancherel’s
theorem which is the result from Fourier’s analysis.
If we have a function
f(x) = 1
√
2π
R +∞
−∞ F(k)eikx
dx –(16)
equation (16) physically means that we are
shifting from x(coordinated space to k space)
F(k) is called the Fourier Transform of f(x).
f(x) is called the inverse Fourier transform of
F(k)
17 Lets solve a problem to see what we have
learnt till now?
Suppose we have a free particle, which is initially con-
fined in the range −a < x < a.The particle is released
at time t=0. At time t = 0 the wave function is as
follows
Ψ(x, 0) = Aif − a < x < a
Ψ(x, 0) = 0 if A and a are constants then find Ψ(x, t)
Thefirst thing we need to do is to normalize Ψ(x, 0) .
This we can do using the conventional method of nor-
malization
R +a
−a |Ψ(x, 0)|2
dx = |A|2
R +a
−a dx = 2a|A|2
= A = − 1
√
2a
—
-(17)
so we have evaluated the constant, which is possible
11

12. only because the free particle is confined in a boundary.
Now we will apply the Plancherel’s theorem and use
equation (15)
φ(k) = 1
√
2a
1
√
2π
R +a
−a eikx
dx = 1
2
√
pia
e−ikx
−ik
|+a
−a = − 1
√
aπ
sin(ka)
k
—
(18)
with this value of φ(k)we can find the solution Ψ(x, t)
as follows
Ψ(x, t) = 1
√
2a
R +∞
−∞
sin(ka)
k
ei(kx−6hk2
2m t)
dk—(19)
In the equation (19) the quantity of significance which
is encountered in many topics of physics is
sin(ka)
a . This equation is perhaps in optics under the
heading of diffraction .
18 What is the significance of this equation and
the quantity involving sine?
The quantity sin(ka)
a is very important in terms of
physics. Mathematically the equation is not very
easy to solve but let us see some limiting cases.
• a very small: If the size of the boundary a is
very small then sin(ka) ≈ ka and
φ(k) =
pa
π—(20)
the equation (20) is flat curve which is due to uncer-
tainty principle. a is position and k is momentum.
If spread in a is large then the spread in k is small.
• If a is very large,the spread in position in large then
12

13. the spread in momentum k is very small. In this
case
φ(k) =
pa
π
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 at x =
±π. The mathematical result is consistent with the
Uncertainty principle which says forlarge a the spread
in k will be small.
19 what is the graphical representation of the
concept explained above?
The diagrammatic representation of the concept dis-
cussed above is given below
In the figure the rectangle shows the variation of prob-
ability density with x at time t = 0 the curve shows the
probability density at time t = ma2
6h
20 what is the physical significance of the quan-
tity Ψk(x, 0)?
The quantity as Ψk(x, 0) as usual has no physical
significance,it is the square of Ψ(x, 0) that has physi-
cal meaning.What we need is information regarding the
particle which cannot be directly obtained from Ψk(x, 0)
which is the wave packet.We need to find 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
different times
14

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

16. wave packet.
21 what is the physical meaning of wave packet?
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 fit inside, we
have to modulate or slightly change the size of
the object . In this case a wave packet is a su-
perposition of sinosodial functions whose ampli-
tude is modulated by a factor φ, the sinosodials
or ripples are contained in an envelope .The sit-
uation looks like as shown in the figure
22 It is better to provide a bit more detailed
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 that the 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 Ψ are :
• • Ψ can interfere with itself
• • Ψ is maximum at the point where the particle is
present
16

17. Figure 3: Figure showing the wave packet
17

18. Figure showing Ψ(x, t) as a superposition of waves
• • Ψ represent single particle or photon it never rep-
resents a large number of particle or photons
Thus if we concentrate on these points we find that the
quantity Ψ contains the behavior of a simple particle or
photon Ψ depends on the factors x and t. Ψ(x, t)isawavethatcarriesin
Ψ(x, t) shows that it is a superposition of harmonic
waves. The two figures below will illustrate this. In
the above figure the Fourier transform of Ψ(x, t) with
average wavelength λ0 in a region of space 4x. On the
other hand the Fourier transform of Ψ(x, t) is shown
below The above figure shows the Fourier transform of
18

19. Figure showing the Fourier Transform of Ψ(x, t) with respect to k
19

20. Ψ(x, t) against wave number k where k = 2π
λ . From
basic mathematical tools we have
4k4x ≥ 1—(21)
returning to the De-Broglie wavelength
4p = 4(h
λ) = h
2π × 4k = h4k–(22)
combining equations (21) and (22) we reach to the un-
certainty relation
4x4p ≥6 h—(23)
23 Well, what about the velocity of the particle
and the velocity of the wave?
Let us see the figure below again
20

21. wave packet containing the simple harmonic oscil-
lations or ripples enclosed in an envelope φ(k)
The velocity of the particle is the velocity of the
envelope and is called group velocity
The velocity of the ripples inside the envelope
and is called group velocity
24 How can we prove that group velocity is the
velocity of the particles,is there any mathe-
matical proof??
Yes there is a specific formalism involving Taylor
expansion but we will use a very simple method . we
have the relation E = hν–(23)
The group velocity is the velocity of the center of the wave and is given by
dν
= dE
=
d( p2
2m)
= p
–(24)
21