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.
- exactly solvable
- Schrödinger equation
- eigen function
- eigen values
1. Potential well
The potential of the system is defined as
The one dimensional Schrödinger equation in Cartesian coordinate is given as
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
The solution of the Eq. (2) is
At , and at , the wave function vanishes since the potential is infinite. Hence, At ,
The addition and subtraction of these equations give two different solutions.
i. . Since , , the energy eigen value is found as
The eigen function is
According to the normalization condition,
Hence the normalized eigen function for is
ii. . For this case,, the corresponding energy eigen value is
The eigen function is and the normalized eigen function is
In Summary, the eigen value is for all positive integer values of “n.” The normalized eigen functions are
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 , which corresponds to the ground state. The ground state energy is
This is known as zero-point energy in the case of the potential well. The excited state energies are , , , and so on. In general, .
The energy difference between the successive states is simply the difference between the energy eigen value of the corresponding state. For example, and . 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 (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 (width of the well is L), the for all positive integers of “n,” the eigen function is and the energy eigen function is .
2. Step potential
Step potential is a problem that has two different finite potentials . 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
The Schrödinger equation in the Region-I and Region-II is given, respectively as,
Case (i): when , the solutions of the Schrödinger equations in the Region-I and Region-II, respectively, are given as
where and . Here, represents the exponentially increasing wave along the x-direction. The wave function must be finite as . This is possible only by setting . Hence the eigen function in the Region-II is
According to admissibility conditions of wave functions, at , and . It gives us
From these two equations,
The reflection coefficient R is given as
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 , since .
Case (ii): when , the solutions are given as
where . As , the wave function must be finite. Hence
by setting . According to the boundary conditions at ,
From these equations,
The reflection coefficient R and the transmission coefficient T, respectively, are given as
From these easily one can show that
The results again indicate that the total number of particles which encounters the step potential is conserved.
3. Potential barrier
In the Region-I, the Schrödinger equation is . The wave function in this region is given as
In Region-II, if , the Schrödinger equation is . The solution of the equation is given as
The Schrödinger equation in the Region-III is . The corresponding solution is . But in the Region-III, the waves can travel only along positive x-direction and there is no particle coming from the right,. Hence
At , and . These give us two equations
At , and . These conditions give us another two equations
The transmission coefficient T is found as
When , 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 , the transmission coefficient is zero. Hence the tunneling is not possible only when .
When the length of the barrier is an integral multiple of , 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) . The potential of the system
Here is the positive constant, which is the strength of the delta potential. Here, we confine ourselves only to the bound states, hence (Figure 4).
The Schrödinger equation is
The solution of the Schrödinger equation is given as
where . At , . So the coefficients and are equal. But , since the first derivative causes the discontinuity. The first derivatives are related by the following equation
This gives us
Equating the value of gives the energy eigen value as
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 gives us . Hence the eigen function is
To normalize ,
This gives us .
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 time-independent Schrödinger equation is given as
The potential is not constant since it is a function of “x”; Eq. (40) cannot solve directly as the previous problems. Let
Using the new constant and the variable , the Schrödinger equation has the form
The asymptotic Schrödinger equation is given as
The general solution of the equation is . As , becomes infinite, hence it cannot be a solution. So the only possible solution is . Based on the asymptotic solution, the general solution of Eq. (42) is given as
The normalized eigen function is
The solution given in Eq. (43) is valid if the condition holds. This gives the energy eigen value as
The important results are given as follows:
The integer represents the ground state, represents the first excited state, and so on. The ground state energy of the linear harmonic oscillator is . This minimum energy is known as ground state energy.
The ground state normalized eigen function is
The energy difference between any two successive levels is . 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 . The Hamiltonian of the linear harmonic oscillator is given as,
Let us define the operator “a,” lowering operator, in such a way that
and the corresponding Hermitian adjoint, raising operator, is
Here, represents the commutation between the operators and . and Eq. (49) becomes
In the same way, one can find the and it is given as
The Hamiltonian H acts on any state that gives the eigen value times the same state , that is,.
The expectation value of is
Let us consider the ground state
Since , . Thus,
Similarly, the energy of the first excited state is found as follows:
In the same way, , , and so on. Hence, one can generalize the result as
The uncertainties in position and momentum, respectively, are given as
a. The expectation value of ‘x’ is given as,
Since the states , , are orthogonal to each other, and . So . The expectation value of the position in any state is zero.
b. The expectation value of momentum is
Not only position, the expectation value of momentum in any state is also zero.
The minimum uncertainty state is the ground state. In this state, and .
Hence the minimum uncertainty product is . Since the other states have higher uncertainty than the ground state, the general uncertainty is . This is the mathematical representation of Heisenberg’s uncertainty relation.
Since corresponds to the low energy state, . This gives us the ground state eigen function. This can be done as follows:
Integrating the above equation gives,
The normalized eigen function is given as
One can see that this result is identical to Eq. (45).
The other eigen states can be evaluated using the equation, .
7. Particle in a 3D box
The three dimensional time-independent Schrödinger equation is given as
Let the eigen functionis taken as the product of , and according to the technique of separation of variables. i.e., .
Divide the above equation by gives us
Now we can boldly write E as
Now the equation can be separated as follows:
The normalized eigen function is given as
In the same way, and are given as
Hence, the eigen function is given as
The energy given values are given as
The total energy E is
Some of the results are summarized here:
In a cubical potential box, , then the energy eigen value becomes,
The minimum energy that corresponds to the ground state is . Here .
Different states with different quantum numbers may have the same energy. This phenomenon is known as degeneracy. For example, the states (i) , (ii) ; and (iii) have the same energy of . So we can say that the energy has 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.