Open access peer-reviewed chapter

Exactly Solvable Problems in Quantum Mechanics

By Lourdhu Bruno Chandrasekar, Kanagasabapathi Gnanasekar and Marimuthu Karunakaran

Submitted: April 29th 2020Reviewed: July 3rd 2020Published: August 18th 2020

DOI: 10.5772/intechopen.93317

Abstract

Some of the problems in quantum mechanics can be exactly solved without any approximation. Some of the exactly solvable problems are discussed in this chapter. Broadly there are two main approaches to solve such problems. They are (i) based on the solution of the Schrödinger equation and (ii) based on operators. The normalized eigen function, eigen values, and the physical significance of some of the selected problems are discussed.

Keywords

• exactly solvable
• Schrödinger equation
• eigen function
• eigen values

1. Potential well

The potential well is the region where the particle is confined in a small region. In general, the potential of the confined region is lower than the surroundings (Figure 1) [1, 2].

The potential of the system is defined as

V=0,L<x<L,Otherwise

The one dimensional Schrödinger equation in Cartesian coordinate is given as

22mΨ+VΨ=EΨΨ+2m2EVΨ=0E1

In the infinite potential well, the confined particle is present in the well region (Region-II) for an infinitely long time. So the solution of the Schrödinger equation in the Region-II and Region-III can be omitted for our discussion right now. The Schrödinger equation in the Region-II is written as

Ψ+2m2EΨ=0
Ψ+α2Ψ=0,whereα2=2mE2E2

The solution of the Eq. (2) is

Ψ=A1sinαx+A2cosαxE3

At x=L, and at x=L, the wave function vanishes since the potential is infinite. Hence, At x=L,

A1sinαL+A2cosαL=0E4

Similarly, atx=L

A1sinαL+A2cosαL=0E5

The addition and subtraction of these equations give two different solutions.

1. i. 2A2cosαL=0cosαL=0αL=/2α2=n2π2/4L2;n=1,3,5,. Since α2=2mE2, 2mE2=n2π2/4L2, the energy eigen value is found as

E=n2π22/8mL2E6

The eigen function is Ψ=A1cosαx

=A1cosnπx/2L

According to the normalization condition,

LLΨΨdx=0A1=L1/2

Hence the normalized eigen function for n=1,3,5,is

Ψ=L1/2cosnπx/2LE7

1. ii. 2A1sinαL=0sinαL=0αL=/2α2=n2π2/4L2;n=2,4,6,. For this case,n=2,4,6,, the corresponding energy eigen value is

E=n2π22/8mL2E8

The eigen function is Ψ=A2cosαxand the normalized eigen function is

Ψ=L1/2sinnπx/2LE9

In Summary, the eigen value is E=n2π22/8mL2for all positive integer values of “n.” The normalized eigen functions are

Ψ=L1/2cosnπx/2L,n=1,3,5,L1/2sinnπx/2L,n=2,4,6,E10

The integer “n” is the quantum number and it denotes the discrete energy states in the quantum well. We can extract some physical information from the eigen solutions.

• The minimum energy state can be calculated by setting n=1, which corresponds to the ground state. The ground state energy is

E1=π22/8mL2E11

This is known as zero-point energy in the case of the potential well. The excited state energies are E2=4π22/8mL2, E3=9π22/8mL2, E4=16π22/8mL2, and so on. In general, En=n2×E1.

• The energy difference between the successive states is simply the difference between the energy eigen value of the corresponding state. For example, E12=E1E2=3E1and E23=E2E3=5E1. Hence the energy difference between any two successive states is not the same.

• Though the eigen functions for odd and even values of “n” are different, the energy eigen value remains the same.

• If the potential well is chosen in the limit 0<x<2L(width of the well is 2L), the energy eigen value is the same as given in Eqs.(6) and (8). But if the limit is chosen as 0<x<L(width of the well is L), the for all positive integers of “n,” the eigen function is Ψ=2/L1/2sinnπx/Land the energy eigen function is E=n2π22/2mL2.

2. Step potential

Step potential is a problem that has two different finite potentials [3]. Classically, the tunneling probability is 1 when the energy of the particle is greater than the height of the barrier. But the result is not true based on wave mechanics (Figure 2).

The potential of the system

V=0,<x<0V0,0x<

The Schrödinger equation in the Region-I and Region-II is given, respectively as,

Ψ+2m2EΨ=0E12
Ψ+2m2EVΨ=0E13

Case (i): when E<V0, the solutions of the Schrödinger equations in the Region-I and Region-II, respectively, are given as

Ψ1=A1expiαx+B1expiαxE14
Ψ2=A2expβx+B2expβx

where α2=2mE2and β2=2mEV02. Here, B2expβxrepresents the exponentially increasing wave along the x-direction. The wave function Ψ2must be finite as x. This is possible only by setting B2=0. Hence the eigen function in the Region-II is

Ψ2=A2expβxE15

According to admissibility conditions of wave functions, at x=0, Ψ1=Ψ2and Ψ1=Ψ2. It gives us

A1+B1=A2E16
A1B1=iβαA2E17

From these two equations,

A2=2αα+A1
B1=αα+A1

The reflection coefficient R is given as

R=B12A12=αα+2=1E18

It is interesting to note that all the particles that encounter the step potential are reflected back. This is due to the fact that the width of the step potential is infinite. The number of particles in the process is conserved, which leads that T=0, since T+R=1.

Case (ii): when E>V0, the solutions are given as

Ψ1=A1expiαx+B1expiαx
Ψ2=A2expiβx+B2expiβx

where β2=2mEV02. As x, the wave function Ψ2must be finite. Hence

Ψ2=A2expiβxby setting B2=0. According to the boundary conditions at x=0,

A1+B1=A2E19
A1B1=βαA2E20

From these equations,

A2=2αα+βA1
B1=αβα+βA1

The reflection coefficient R and the transmission coefficient T, respectively, are given as

R=B12A12=αβα+β2E21
T=A22A12=4αβα+β2E22

From these easily one can show that

T+R=4αβα+β2+αβα+β2=1E23

The results again indicate that the total number of particles which encounters the step potential is conserved.

3. Potential barrier

This problem clearly explains the wave-mechanical tunneling [3, 4]. The potential of the system is given as (Figure 3)

V=V0,0<x<L0,Otherwise

In the Region-I, the Schrödinger equation is Ψ+α2Ψ=0. The wave function in this region is given as

Ψ1=A1expiαx+B1expiαxwhereα2=2mE2E24

In Region-II, if E<V0, the Schrödinger equation is Ψβ2Ψ=0. The solution of the equation is given as

Ψ2=A2expβx+B2expβxwhereβ2=2mEV02E25

The Schrödinger equation in the Region-III is Ψ+α2Ψ=0. The corresponding solution is Ψ3=A3expiαx+B3expiαx. But in the Region-III, the waves can travel only along positive x-direction and there is no particle coming from the right,B3=0. Hence

Ψ3=A3expiαxE26

At x=0, Ψ1=Ψ2and Ψ1=Ψ2. These give us two equations

A1+B1=A2+B2E27
A1B1=βA2B2E28

At x=L, Ψ2=Ψ3and Ψ2=Ψ3. These conditions give us another two equations

A2expβL+B2expβL=A3expiαLE29
A2expβLB2expβL=A3βexpiαLE30

Solving the equations from (27) to (30), one can find the coefficients in the equations. The reflection coefficient is R is found as

R=B12A12=V024EV0Esinh2βL1+V024EV0Esinh2βL1E31

The transmission coefficient T is found as

T=A22A12=1+V024EV0Esinh2βL1E32

From Eqs. (31) and (32), one can show that T+R=1. The following are the conclusions obtained from the above mathematical analysis.

• When E<V0, though the energy of the incident particles is less than the height of the barrier, the particle can tunnel into the barrier region. This is in contrast to the laws of classical physics. This is known as the tunnel effect.

• As V0, the transmission coefficient is zero. Hence the tunneling is not possible only when V0.

• When the length of the barrier is an integral multiple of π/β, there is no reflection from the barrier. This is termed as resonance scattering.

• The tunneling probability depends on the height and width of the barrier.

• Later, Kronig and Penney extended this idea to explain the motion of a charge carrier in a periodic potential which is nothing but the one-dimensional lattices.

4. Delta potential

The Dirac delta potential is infinitesimally narrow potential only at some point (generally at the origin, for convenience) [3]. The potential of the system

V=λδx,x=00,Otherwise

Here λis the positive constant, which is the strength of the delta potential. Here, we confine ourselves only to the bound states, hence E<0(Figure 4).

The Schrödinger equation is

Ψ+2m2EVΨ=0Ψ+2m2E+λδxΨ=0E33

The solution of the Schrödinger equation is given as

RegionI:Ψ1=A1expβxE34
RegionII:Ψ2=A2expβxE35

where β2=2mE2. At x=0, Ψ1=Ψ2. So the coefficients A1and A2are equal. But Ψ1Ψ2, since the first derivative causes the discontinuity. The first derivatives are related by the following equation

Ψ2Ψ1=22E36

This gives us

β=2E37

Equating the value of βgives the energy eigen value as

E=mλ222E38

The energy eigen value expression does not have any integer like in the case of the potential well. Hence there is only one bound state which is available for a particular value of “m.”

The eigen function can be evaluated as follows: The eigen function is always continuous. At x=0gives us A1=A2=A. Hence the eigen function is

Ψ=Aexpβx

To normalize Ψ,

Ψ2dx=120Ψ2dx=1

This gives us A=β=.

5. Linear harmonic oscillator

Simple harmonic oscillator, damped harmonic oscillator, and force harmonic oscillator are the few famous problems in classical physics. But if one looks into the atomic world, the atoms are vibrating even at 0 K. Such atomic oscillations need the tool of quantum physics to understand its nature. In all the previous examples, the potential is constant in any particular region. But in this case, the potential is a function of the position coordinate “x.”

5.1 Schrodinger method

The potential of the linear harmonic oscillator as a function of “x” is given as (Figure 5) [4, 5, 6]:

V=2x22E39

The time-independent Schrödinger equation is given as

Ψ+2m2E2x22Ψ=0E40

The potential is not constant since it is a function of “x”; Eq. (40) cannot solve directly as the previous problems. Let

α=1/2x and β=2Eω.

Using the new constant βand the variable α, the Schrödinger equation has the form

d2Ψdα2+βα2Ψ=0E41

The asymptotic Schrödinger equation αis given as

d2Ψdα2α2Ψ=0E42

The general solution of the equation is exp±a2/2. As α, exp+a2/2becomes infinite, hence it cannot be a solution. So the only possible solution is expa2/2. Based on the asymptotic solution, the general solution of Eq. (42) is given as

Ψ=Hnαexpa2/2

The normalized eigen function is

Ψ=π1/212n×n!1/2Hnαexpa2/2E43

The solution given in Eq. (43) is valid if the condition 2n+12Eω=0holds. This gives the energy eigen value as

E=n+12ωE44

The important results are given as follows:

• The integer n=0represents the ground state, n=1represents the first excited state, and so on. The ground state energy of the linear harmonic oscillator is E=ω/2. This minimum energy is known as ground state energy.

• The ground state normalized eigen function is

Ψ0x=π1/4expx22E45

• The energy difference between any two successive levels is ω. Hence the energy difference between any two successive levels is constant. But this is not true in the case of real oscillators.

5.2 Operator method

The operator method is also one of the convenient methods to solve the exactly solvable problem as well as approximation methods in quantum mechanics [5]. The Hamiltonian of the linear harmonic oscillator is given as,

H=p22m+12mω2x2E46

Let us define the operator “a,” lowering operator, in such a way that

a=21/2mωx+ipE47

and the corresponding Hermitian adjoint, raising operator, is

a+=21/2mωxipE48
a+a=2mωℏ1mωxipmωx+ip=2mωℏ1m2ω2x2+p2+imωxpimωpx=21m2ω2x2+p2+imωxpE49

Here, xprepresents the commutation between the operators xand p. xp=iℏand Eq. (49) becomes

a+a=2mωℏ1m2ω2x2+p2mωℏ
=1ωℏ12mω2x2+p22m12
=Hω12E50

In the same way, one can find the aa+and it is given as

aa+=Hω+12E51

Adding Eqs. (50) and (51) gives us the Hamiltonian in terms of the operators.

H=ω2aa++a+aE52

Subtracting Eq. (50) from (51) gives, aa+a+a=1. This can be simplified as

aa+=1E53

The Hamiltonian H acts on any state nthat gives the eigen value Entimes the same state n, that is,Hn=Enn.

The expectation value of a+ais

a+ana+an=nHℏω12n=1ℏωnHnn12n=1ωEnnn12=Enω12E54

Let us consider the ground state 0.

0a+a0=E0ℏω12

Since a0=0, 0a+a0=0. Thus,

E0ω12=0E0=ω2E55

Similarly, the energy of the first excited state is found as follows:

1a+a1=E1ℏω12
11a+0=E1ℏω12
1.111=E1ℏω12
1=E1ω12E1=32ωE56

In the same way, E2=5ω/2, E3=7ω/2, and so on. Hence, one can generalize the result as

En=n+12ωE57

The uncertainties in position and momentum, respectively, are given as

x=x2x2E58
p=p2p2E59

In order to evaluate the uncertainties x2, x2, p2, and p2have to be evaluated. From Eqs. (47) and (48) the position and momentum operators are found as

x=21/2a+a+E60
p=21/2aa+iE61

1. a. The expectation value of ‘x’ is given as,

xnxn=21/2na+a+n=21/2nan+na+n=21/2nnn1+n+1nn+1

Since the states n1, n, n+1are orthogonal to each other, nn1=0and nn+1=0. So x=0. The expectation value of the position in any state is zero.

1. b. The expectation value of momentum is

pnpn=21/21inaa+np=0.

Not only position, the expectation value of momentum in any state is also zero.

1. c.

x2nx2n=2na+a+a+a+n=2na2+a+2+aa++a+an=2na2n+na+2n+naa+n+na+an=2nn1nn2+n+1n+2nn+2+n+1nn+nnn=22n+1

1. d.

p2np2n=2naa+aa+n=mωℏ2na2n+na+2nnaa+nna+an=mωℏ2nn1nn2+n+1n+2nn+2n+1nnnnn=mωℏ22n+1

From Eq. (58) and (59), the uncertainty in position and momentum, respectively are given as,

x=22n+11/2E62
p=22n+11/2E63
x.p=22n+1E64

6. Conclusions

• The minimum uncertainty state is the ground state. In this state, x=21/2and p=mωℏ21/2.

• Hence the minimum uncertainty product is x.p=2. Since the other states have higher uncertainty than the ground state, the general uncertainty is x.p2. This is the mathematical representation of Heisenberg’s uncertainty relation.

• Since Ψ0xcorresponds to the low energy state, aΨ0x=0. This gives us the ground state eigen function. This can be done as follows:

aΨ0x=0
21/2mωx+ipΨ0x=0
21/2x+iiℏ∂/x21/2Ψ0x=0
Ψ0xx=xΨ0x
dΨ0xΨ0x=mωxdx

Integrating the above equation gives,

lnΨ0x=x22+lnA
Ψ0x=Aexpx22

The normalized eigen function is given as

Ψ0x=π1/4expx22

One can see that this result is identical to Eq. (45).

• The other eigen states can be evaluated using the equation, Ψnx=a+n/n!Ψ0x.

7. Particle in a 3D box

The confinement of a particle in a three-dimensional potential is discussed in this section [4, 6]. The potential is defined as (Figure 6)

V=0,0x<a;0y<b;0z<c,Otherwise

The three dimensional time-independent Schrödinger equation is given as

2Ψxyz2m2VΨxyz=xyzE65

Let the eigen functionΨxyzis taken as the product of Ψxx, Ψyyand Ψzzaccording to the technique of separation of variables. i.e., Ψxyz=ΨxxΨyyΨzz.

ΨyyΨzzd2Ψxxdx2+ΨxxΨzzd2Ψyydy2+ΨxxΨyyd2Ψzzdz22m2VΨxyz=2m2xyz

Divide the above equation by Ψxyzgives us

1Ψxxd2Ψxxdx2+1Ψyyd2Ψyydy2+1Ψzzd2Ψzzdz2=2m2EE66

Now we can boldly write E as Exx+Eyy+Ezz

1Ψxxd2Ψxxdx2+1Ψyyd2Ψyydy2+1Ψzzd2Ψzzdz2=2m2Exx+Eyy+EzzE67

Now the equation can be separated as follows:

d2Ψxxdx2+2m2ExxΨxx=0
d2Ψyydy2+2m2EyyΨyy=0
d2Ψzzdz2+2m2EzzΨzz=0

The normalized eigen function Ψxxis given as

Ψxx=2a1/2sinnxπxa

In the same way, Ψyyand Ψzzare given as

Ψyy=2b1/2sinnyπyb
Ψzz=2c1/2sinnzπzc

Hence, the eigen function Ψxyzis given as

Ψxyz=ΨxxΨyyΨzz=8abc1/2sinnxπxasinnyπybsinnzπzcE68

The energy given values are given as

Exx=nx2π222ma2
Eyy=ny2π222mb2
Ezz=nz2π222mc2

The total energy E is

E=Exx+Eyy+Ezz=π222mnx2a2+ny2b2+nz2c2E69

Some of the results are summarized here:

• In a cubical potential box, a=b=c, then the energy eigen value becomes,

E=π222ma2nx2+ny2+nz2.

• The minimum energy that corresponds to the ground state is E1=3π222ma2. Here nx=ny=nz=1.

• Different states with different quantum numbers may have the same energy. This phenomenon is known as degeneracy. For example, the states (i) nx=2;ny=nz=1, (ii) ny=2;nx=nz=1; and (iii) nz=2;nx=ny=1have the same energy of E=6π22ma2. So we can say that the energy 6π22ma2has a 3-fold degenerate.

• The states (111), (222), (333), (444),…. has no degeneracy.

• In this problem, the state may have zero-fold degeneracy, 3-fold degeneracy or 6-fold degeneracy.

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Lourdhu Bruno Chandrasekar, Kanagasabapathi Gnanasekar and Marimuthu Karunakaran (August 18th 2020). Exactly Solvable Problems in Quantum Mechanics, Quantum Mechanics, Paul Bracken, IntechOpen, DOI: 10.5772/intechopen.93317. Available from:

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