One Dimensional System | Modern Physics PDF Download

Properties of one-dimensional motion:

(A) Bound states (quantum mechanical discrete spectrum)
Bound states occur whenever the particle cannot move to infinity and particle is confined into limited region. 

• From the figure the condition for bound states is V_min < E < V₁ 
• In a one dimensional potential energy, level of a bound state system are discrete and non degenerate. 
• The wave function Ψ_n of one dimensional bound state system has n nodes i.e., Ψn vanishes n times, if n corresponds to n = 0 and n - 1 nodes, if n = 1 corresponds to the ground state.

(B) Continuous spectrum (unbound states) 
Unbound states occur in those cases where the motion of the system is not confined in above figure. 

• V₁ < E < V₂ : The energy spectrum is continuous and none of the energyeigen values is degenerate. 
• E > V₂ : The energy spectrum is continuous and particles motion is infinite in both directions. And this spectrum is doubly degenerate.

Current Density (J)
The probability current density is defined as,

The current density can also be given as,
J = ρv
where ρ is probability density. i.e., ρ = |ψ|² and v is velocity of particle which is ℏk/m in momentum space.
So, the current density is

The current density and probability density will satisfy the continuity equation which is given by,

Free Particle in One Dimension 

The potential of free particle is defined as, V (x) =  0; -∞ < x < ∞
The Schrödinger wave function is given by
H = Eψ

The solution is given by,

The energy eigen value of free particle is  is continuous and wave function is ψ₊ = Ae^ikx and ψ_- = Ae^-ikx, where ψ₊ moves from positive x -axis and ψ_- moves from negative x -axis.
Hence, there are two eigenfunctions for energy,  then wave function is doubly degenerate. 

The Step Potential

If J_i is incident current density, J_r is reflection current density and J_t is transmission current density, then reflection coefficient is defined as  and transmission coefficient is defined as
The potential step is defined as, 

Case I: E > V  
For x < 0 , the Schrödinger wave equation is given as

where  is incoming wave andis reflected wave. In region x > 0 , where the Schrödinger wave equation is-

D = 0  as no wave reflected in region II  i.e., x > 0
 which is transmitted wave i.e.,

Using Boundary condition at x = 0 i.e.,
Wave must be continuous and differentiable at boundary.
So, ψ₁(x = 0) = ψ₂(x = 0) ⇒ A + B = C

 Solution of above two equations are,

Case II: E < V₀ Schrodinger wave equation for x > 0
Hψ = Eψ

ψ = Ae^ikx + Be^ikx 
If  is incoming wave,  then  is reflected wave,
Schrödinger wave equation for x > 0

A = 0 , as wave function must vanish at x → ∞ , then

Now, we apply boundary condition at x = 0 i.e.,
ψ₁ (x = 0) = ψ₂ (x = 0) ⇒ (A + B) = C

when E < V₀ , there is a finite probability to find the particle at x > 0 , even if E < 0 but current density is zero in region x > 0.
Strange part of the problem is that, even if transmission coefficient is zero there is finite probability to find the particle at x > 0.

Particle in a One Dimensional Box

The potential of one dimensional box is defined as

Time independent Schrödinger wave equation is given as, Hψ = Eψ for region x < 0 and x > 0 , ψ(x) = 0, because in this region potential is infinity, so probability to find the particle in that region is zero.
Schrodinger wave equation in the region, 0 < x < a
Hψ = Eψ

Now wave function must be continuous at the boundary
So, ψ(0) = ψ(a) = 0 ⇒ 0 = A sin0 + B cos0 = B ⇒ B = 0
Therefore, ψ(x) = A sin kx
Now, ψ(a) = 0 ⇒ A sin ka = 0 ⇒ ka = nπ, where n = 0, 1, 2, 3, ....
But for n = 0 ψ(x) = 0 so n = 0 is not possible
So,  n = 1, 2, 3, .... , ka = nπ

E_n is energy eigen value which is discrete.
 the value of A can be find with normalization condition which is

So, energy eigen function is given by
Energy eigen value is given as
The orthonormal condition is given by

Any function f(x) can be expressed in the term of ψ_n (x).

Infinite Symmetric Potential Box 

Wave function in region,  vanishes i.e., ψ(x) = 0, because potential is infinite in this region. In region,  the solution of Schrödinger wave function is given by, ψ = A sin kx + B cos kx
Parity operator will commute with Hamiltonian because wave function can have either even or odd symmetry.
Now wave function must vanish at boundary. 

Solution for infinite symmetric box as mentioned above is given by  

and energy eigen value,

First three eigenstate are shown in the above figure. 

Square Well Finite Potential Box (graphical method) 

Square well finite potential box is defined as,

For Bound state, E < 0
Schrödinger wave solution in region I, i.e.,
ϕ₁(x) = Ae^γx + Be-^γx 
The wave function must vanish at x → -∞ , i.e., wave function must be zero.
So, B = 0
Thus, ϕ₁(x) = Ae^γx ;
Schrodinger wave solution in region II, i.e.,

ϕ_I(x) = B cos kx;  for even parity
ϕ_II(x) = B cos kx; for odd parity
(potential is symmetric about x = 0. So parity must be conserved).
Where, k² = 2mE/ℏ² 
Schrodinger wave solution  in region III i.e., x > a/2
ψ =  De^-γx + Ee^γx 
The wave function must vanish at x → ∞ so E = 0

In the table below shown the number of bound states for various range of V₀a². where R denotes the radius.

Harmonic Oscillator (Parabolic potential) 

The parabolic potential is defined as,

The Schrödinger wave function is given as, 

The general solution is given by

So, equation (i) reduces to
Solving series solution by putting

When k = 2n + 1, the equation (i) reduces to Hermite polynomial and

Hermite polynomials and it is given as

Energy eigen function is

And eigen value is
Ground state n = 0 = E = ℏω/2 is zero point energy. 

The first three stationary state and corresponding eigen value for the harmonic oscillator. 

Problems on One Dimensional System: 
Example: A particle of mass m is confined into a box of width L , where potential is defined as,  

Find
(a) 〈x〉
(b) 〈x²〉
(c) 〈P_x〉
(d) 〈P_x²〉
(e) Δx · ΔP_x 

Example: A particle of mass m which moves freely inside an infinite potential well of length a has the following initial wave function.

(a) Find A so that ψ(x, 0) is normalized.
(b) If measurement of energy is carried out, what are the values that will be found and   what are the corresponding probability?
(c) Calculate the average energy.

Particle of mass m confine into a box of width a.

For normalized Ψ,

If energy will be measured on state |Ψ〉, the measurement of |Ψ〉, yields either  which is eigen value associated with |ϕ₁〉, |ϕ₃〉, |ϕ₅〉 respectively.

(c) Average energy is given by 〈E〉 = ∑E_i P_i 

Example: Prove for any normalized wave function of particle of mass m in one dimensional,

Example: Three dimensional wave function is given by, Ψ(r) = (A/r)e^ikr 
Find the current density.
Ψ(r) = (A/r)e^ikr 

Example: A potential barrier is given as
Prove that the expression of transmission probability for E < V₀ is given as,

The potential is given as
 

Case - E < V₀ 
The Schrödinger wave solution in region I, II and III is given by -

Boundary condition -

The transmission probability is given by

Solving the above four boundary condition

For approximation V₀ > E,

Example: Consider a particle of mass m and charge q placed in uniform field,Apart from this, a restoring force corresponding to the potential V(x) = 1/2 mω² X² acts.
Find the lowest energy eigen value. Consider the electric field is originated at origin.

The electric potential energy at the position x will be -qEX. So the effective potential is given by

So, Hamiltonion is given by-


So, Energy is given by

Example: A particle of energy 9eV is sent towards a potential step 8eV high.
(a) What is reflection coefficient.
(b) What percentage will transmitted.

Example: A particle in the infinite square well has the initial wave function.
ψ(x, 0) = Ax(a - x); (0 ≤ x ≤ a)
(a) Find the value of A such that ψ is normalized.
(b) Write down ψ (x) in the basis of ϕ_n (x), where ϕ_n (x) is the eigen function of the n^th state (wave function) for the system confined into box whose potential is given as,

(c) Write down expression of ψ (x, t)

(a) ψ(x, 0) = Ax(a- x); 0 ≤ x ≤ a


And c_n can be found with Fourier’s trick



From Schrodinger wave equation,

Example: |ϕ_n〉 represent the energy eigen state of a linear harmonic oscillator and if state  of harmonic state of angular frequency ω.
(a) If energy is measured what will measurement with what probability
(b) Find average value of energy.

(a) If energy is measured, the measurement is ℏω/2 and 3ℏω/2 with probability 1/2 and 1/2