As discussed in the article of time dependent Schrodinger wave equation:

V=A exp[-i/Ћ(Et-px]

= A exp(-i/Ћ Et) exp(i/Ћ)

Ψ=ψ^{’} exp(-iEt/Ћ) (1)

Where Ћ = h/2π

So, ψ is a product of a time dependent function exp(-i/Ћ Et) and a position dependent function

Ψ^{’}= A exp(-i/Ћ px)

Differentiating equation (1) w.r.t.x, We have

dψ/dx = exp(-i/ЋEt) dψ^{’}/dx

and d^{2}ψ/dx^{2}= exp(-i/Ћ Et) d^{2}ψ^{’}/dx^{2} (2)

Also on differentiating ψ w.r.t. t, we have

dψ/dt=ψ^{’ }exp (-iEt/Ћ) (I E/Ћ)

dψ/dt=-(iE/Ћ)ψ^{’} exp(-I Et/Ћ) (3)

Put equations [1-3] in time dependent Schrodinger wave equation (discussed earlier),

iЋ[-iE/Ћψ^{’} exp(-iEt/Ћ)]= -Ћ^{2}/2m[exp(i/ЋEt) d^{2}ψ^{’}/dx^{2}] +V ψ^{’ }exp(iEt/Ћ)

Eψ^{’} exp(iEt/Ћ) = -Ћ^{2}/2m exp(i/Ћ Et) d^{2}ψ^{’}/dx^{2} + V ψ^{’}exp(iEt/Ћ)

Dividing throughout by expression (i/Ћ Et) we have

Eψ^{’}= (-Ћ^{2}/2π) d^{2}ψ^{’}/dx^{2}+V ψ^{’}

Or (E-V)ψ^{’}=-Ћ^{2}/2m dψ^{’}/dx^{2}

or d^{2}Ψ^{’}/dx^{2} + (2m/Ћ^{2})(E-V)ψ^{’} (4)

Which is time independent form of Schrodinger wave equation in one dimension.

In three-dimensional form:

d^{2} Ψ^{’}/dx^{2}+ d^{2}ψ^{’}/dy^{2}+ d^{2}Ψ’/ d^{2}x^{2}+2m/Ћ^{2}(E-V)ψ^{’}=0

In this equation, ψ’ equation, ψ’(x) is also called the wave function. The potential V(x) does not contain the time explicity and E, the total energy of the particle is a constant.

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dear sir,

thx for providing this article on winnerscience….

yours obediently,

raghav sharma

b.tech (ece,1)

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