Particle in a Potential Well Vc[1]

8/3/2019 Particle in a Potential Well Vc[1]

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Particle in a Potential Well

Figure 1

Figure 1 represents an infinite square well potential of width a. The potential

of this one dimensional well is V(x) = 0 for 0 x a, and V(x) = for x a.

1

V(x)

x = 0 x = ax


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The infinite square well potential corresponds to the motion of a particle

constrained by impenetrable walls to move in a region of width a where the

potential energy is constant. The potential energy is taken to be zero as the

potential energy is infinite at x = 0 and x = a, the probability of finding the

particle outside the well is zero. The wave function (x) thus vanishes for

xa.

However, there is a wave function in the region 0 x a

2


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Recall that the one dimensional time independent Schrodinger equation is

2 2

2( )

2

dV x E

m dx

+ =

h

Since V(x) = zero in this region, the second term vanishes and we now have

2 2

20

2

dE

m dx

+ =

h

or2

2 2

20

d mE

dx

+ =

h.

The Schrodinger equation of the particle with mass m and energy E in the

region 0 x a can thus be expressed as2

2 2

20

d mE

dx

+ =

h.

3


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Remember that the wave function must be continuous; then (x) must

vanish at the walls. The general solution of the Schrodinger equation is

1 22 2

2 2( ) cos sin

mE mE x c x c x

= +

h h

But (0) = 0

(0) = 0

( ) ( ) ( )1 2

1

0 cos 0 sin 0 0

0

c c

c

= + =

=

Thus the wave function can be expressed as

( ) 2 22

sinmE

x c x

= h

4


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Since the second boundary condition is (a) = 0, we have

( ) 2 22

sin 0mE

a c a

= = h

But c2 0 for the wave function to be in existence.

2

2

2sin 0

2or where 1, 2,3...

mEa

mEa n n

=

= =

h

h

Note that we exclude the case n = 0, which would also produce a wave

function that vanishes everywhere. This is contrary to the initial conditions

where we know that a wave function exists in a certain region i.e. the region

where 0 x a Since a is also fixed, the only free parameter is the energy

E. This tells us that the Schrodinger equation has a solution only for

certain discrete values ofE.

2 2

2

2, where 1, 2, 3...

2 E n n

ma= =h

5


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Note also that the smallest possible energy (n = 1) for the particle is not zero but

2 2

1 22E

ma= h

This is the minimum energy and is called the zero point energy of the infinite

potential well with Edefined, we now go back to the wave function

( ) 2 22

sinmE

x c x

= h

( )

( )

2

2

sin

or sinn

n xx c

a

n xx c

a

=

=

The constant c2 can be obtained by normalizing the wave function

( ) ( )2 2

2

2

2

2

sin

2

1

2

a a

n no o

n x x x dx c dxa

ac

ca

=

=

=

=

6


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Aside

2 2

sin sin

a a

o o

n x a n x n

dx d xa n a a

=

Let

0, 0

,

n x

a

x

x a n

=

= == =

( )

2 2

sin sin

11 cos 2

2

1 cos 2

2 2

sin 2

2 4

sin2

2 4

2

a n

o o

n

o

n n

o o

n n

o o

n x a

dx da n

ad

n

ad d

n

a

n

a n n

n

a

=

=

=

=

=

=

( )

( )

2sin for 0

and 0 for

n

n

n x x x a

a a

x x x a

=

=

7


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The first 3 stationary states are shown below

The wave functions n (x) are alternatively symmetric (for odd n) and antisymmetric (for even n) about x = 2

a .

Note that the solutions of the Schrodinger equation will be either symmetric or

anti symmetric.

8

1 (x)

n = 1

o a

x

2 (x)

n = 2

o ax

3

(x)

o ax


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The function n (x) are mutually orthogonal, in the sense that

( ) ( ) 0m n x x dx =

whenever m n

( ) ( )2

sin sin

1cos cos

a

m no

a

o

m x n x x x dx dx

a a a

m n m n x x dx

a a a

=

+ =

Aside : ( ) ( )1

sin sin cos cos2

A B A B A B= +

( ) ( )( ) ( )

( ) ( )

( )

1 1sin sin

sin sin

0

0

a

m n

o

m n m n x x dx x x

m n a m n a

m n m n

m n m n

+ = +

+ = +

=

=

9


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THE UNCERTAINTY PRINCIPLE

The Uncertainty Principle states that the more precisely determined a particles

position, the less precisely the momentum is. The Heisenberg position

momentum uncertainty principle can thus be expressed as

xx p h

10


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Consider a particle of mass m in a one dimensional square well of width L. Thus

x L and p L

hif v= 0 at the bottom of the well, the energy of the

particle

2 2 2

2 2 22 2 8p hE

m mL mL= =h

If we apply the result to an electron trapped within L 1 , we have

10 E eV

This is about right for the K.E of an electron in the ground state of a hydrogen atom.

Thus the uncertainty principle gives a direct and simple approach to the estimation

of ground state energies.

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Particle in a Potential Well Vc[1]

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