Wave particle duality, Schrodinger time independent wave equation and wave fucntion and probability

Wave particle duality

Like radiation, matter is also associated with wave nature and particle nature. In 1901, Max Plank showed the relation, E = hν. It justify the wave nature of particle or matter. In 1905, Albert Einstein gave the relation, E = mc². It showed the particle nature of matter. On the basis fo these two equations, de - Broglie deduced a relation in 1923 which is termed as de - Broglie wave equation. That equation is x = \(\frac{h}{mv}\) = \(\frac{h}{p}\)

Deduction of de - Broglie wave equation

For a paticle showing wave nature E = hν .......(i)

Here, E = energy

h = plank's constant, h = 6.62× 10^-34Js

ν = frequency of radiation

we know, c =νλ

λ = wavelength

ν = \(\frac{c}{\lambda}\)

Now, equation (i) becomes, E = \(\frac{hc}{\lambda}\) .......(i)

From the mass-energy equivalence relation, E = mc².....(ii)

From equation (ii) and (iii)

mc²=\(\frac{hc}{\lambda}\)

or,λ = \(\frac{h}{mc}\) ......(iv)

For a particle of velocity v, we replace c by v in equation (iv)

λ = \(\frac{h}{mv}\).......(v)

Equation (v) shows the wave nature and particle nature of matter. So the wavelength 'λ' is also called matter wave.

We know, momentum (p) = mass(m)× velocity(v)

Now, equation (v) becomes

λ = \(\frac{h}{p}\)......(vi)

λ = \(\frac{h}{\sqrt 2mKE}\)

Q. An electron is moving with 10⁶m/sec. Calculate its wavelength.

Mass of an electron = 9.1× 10^-31kg

λ = \(\frac{h}{mv}\)

= \(\frac{6.626 \times10^-34}{9.1 \times10^-31 \times10^6}\)

= 0.73× 10^{-9 m}

= 7.3× 10^-8m

Schrodinger time independent wave equation

When matter passes wave, then this should be explained in terms of wave equation. Hence, matter can be explained in terms of stationary wave equation. The wave function for such is given by

Ψ = A sin\(\frac{2πx}{\lambda}\).....(i)

Differentiating the equation (i) with respect to x

\(\frac{d\psi}{dx}\) = A \(\frac{d}{dx}\) sin \(\frac{2\pi x}{\lambda}\)

\(\frac{d\psi}{dx}\) = A \(\frac{2\pi}{\lambda}\) cos \(\frac{2\pi x}{\lambda}\)

Again, differentiating the equation with respect to x

\(\frac{d}{dx}\)(\(\frac{d\psi}{dx}\)) = \(\frac{2\pi A}{\lambda}\)\(\frac{d}{dx}\) cos \(\frac{2\pi x}{\lambda}\)

or, \(\frac{d^2\psi}{dx^2}\) =\(\frac{2\pi A}{\lambda}\) (-\(\frac{2\pi}{\lambda}\)) sin \(\frac{2\pi x}{\lambda}\)

= - \(\frac{4\pi^2}{\lambda^2}\) A sin \(\frac{2\pi x}{\lambda}\)

\(\frac{d^2\psi}{dx^2}\) = - \(\frac{4\pi^2}{\lambda^2}\) \(\psi\)....(ii)

From the de-Broglie wave equatjion, \(\lambda\) = \(\frac{h}{mv}\)

\(\lambda^2\) = \(\frac{h^2}{m^2v^2}\)

= \(\frac{h^2}{2m\frac{1}{2}mv^2}\)

= \(\frac{h^2}{2mK.E}\)

=\(\frac{h^2}{2m(E-V)}\)

We have,

Total energy of particle (E) = K.E + P.E(v)

∴K.E = E-V

Substituting the value of \(\lambda^2\) in equation (ii)

\(\frac{d^2\psi}{dx^2}\) = -\(\frac{4\pi^2 2m(E-V)\psi}{h^2}\)

\(\frac{d^2\psi}{dx^2}\) = -\(\frac{8\pi^2 m(E-V)\psi}{h^2}\)

\(\frac{d^2\psi}{dx^2}\) +\(\frac{8\pi^2 m(E-V)\psi}{h^2}\) = 0....(iii)

This is the Schrodinger wave equation for a particle having mass(m), total energy (E), potential energy (v) along the direction x. The Schrondinger wave equation for the particle along x-axis, y-axis and z-axis is

\(\frac{d^2\psi}{dx^2}\) +\(\frac{d^2\psi}{dy^2}\) +\(\frac{d^2\psi}{dz^2}\) +\(\frac{8\pi^2 m(E-V)\psi}{h^2}\) = 0......(iv)

Schrodinger wave equation in terms of Laplacian operater

\(\bigtriangledown^2\) \(\psi\)+\(\frac{8\pi^2 m(E-V)\psi}{h^2}\) = 0......(v)

Here,\(\bigtriangledown^2\) is called Laplacian operator

\(\bigtriangledown^2\) =\(\frac{d^2}{dx^2}\) +\(\frac{d^2}{dy^2}\) +\(\frac{d^2}{dz^2}\)

Schrodinger wave equation in terms of Hamiltonian operator, multiplying by \(\frac{h^2}{8\pi^2 m}\) for the equation (v)

\(\frac{h^2}{8\pi^2 m}\)\(\bigtriangledown^2\)\(\psi\) + (E-V) \(\psi\) = 0

\(\frac{h^2}{8\pi^2 m}\)\(\bigtriangledown^2\)\(\psi\) + E\(\psi\) - V\(\psi\) = 0

\(\frac{h^2}{8\pi^2 m}\)\(\bigtriangledown^2\)\(\psi\) -V\(\psi\) =-E\(\psi\)

-\(\frac{h^2}{8\pi^2 m}\)\(\bigtriangledown^2\)\(\psi\) +V\(\psi\) =E\(\psi\)

-\(\frac{h^2}{8\pi^2 m}\)\(\bigtriangledown^2\) + v] \(\psi\) =E\(\psi\)

H\(\psi\) =E\(\psi\)....(vi)

Here, H is called Hamiltonian operator

H =-\(\frac{h^2}{8\pi^2 m}\)\(\bigtriangledown^2\) + v]

Wave function and probability

Wave function is a mathematical function which is the measure of amplitude in the propagation of wave. It may have positive or negative or imaginary value. Only the \(\psi\) has no physical significance but \(\psi\)^*.\(\psi\) has physical significance. Here, \(\psi\)^*is a complex conjugate of wavefunction \(\psi\). The complex conjugate is made by changing the sign of i.

Example: \(\psi\) = 2+ i3 then \(\psi\)^*= 2-i3

\(\psi\) =2-i3 then \(\psi\)^*= 2+ i3

\(\psi\) =2 then \(\psi\)^*= 2

\(\psi\) = -i4 then \(\psi\)^*= i4

\(\psi\) = (a + ib) then \(\psi\)^*= a - ib

If \(\psi\) propagate along x-axis it is denoted by \(\psi\) (x). Then its complex conjugate is\(\psi\)^*(x). The probability of finding a particle is obtained by multiplying a wave function with its complex conjugate.

Probability = \(\psi\)^*(x) \(\psi\) (x) along axis

Probability =\(\psi\)^*(x, y, z). \(\psi\) (x, y, z) dx. dy. dz

= \(\psi\)^*.\(\psi\) (d\(\tau\) (d\(\tau\) = dx. dy. dz)

Probability density = \(\frac{probability}{volume element}\)

=\(\frac{\psi^* \psi d \tau}{d \tau}\)

= \(\psi\)^*.\(\psi\)

The probability of finding a particle within the volume element dx.dy.dz or d\(\tau\) is

\(\psi\)^*(x, y, z). \(\psi\) (x, y, z) dx. dy. dz

or,\(\psi\)^*.\(\psi\)d\(\tau\)

or,⟨\(\psi\) /\(\psi\)⟩

The probability density = \(\frac{probability}{volume element}\)

= \(\frac{\psi^* \psi d \tau}{d \tau}\)

= \(\psi\)* \(\psi\)

Here, the term probability density is mentioned because the exact position of microscopic particle cannot be determined. This is in accordance with the Heisenberg Uncertaninty Principle.

Physical significance of wave function \(\psi\)

The wave function \(\psi\) is the measure of amplitude of wave. |\(\psi\)| or \(\psi\)* \(\psi\) has significnace as it is the measure of probability density.

1. \(\psi\) must be finite.
2. \(\psi\) must be single valued.
3. \(\psi\) must be continuous.
4. \(\psi\) must be normalised i,e\(\psi\)^*.\(\psi\)d\(\tau\) = 1
5. \(\psi\) must be real.

References

wikipedia. n.d. <https://en.wikipedia.org/wiki/Wave–particle_duality>.

Jha, V.K. Introductory Quantum Mechanics. Kathmandu, 2012.

vergil.chemistry.gatech.edu. n.d. <http://vergil.chemistry.gatech.edu/notes/quantrev/node8.html>.