Development of Schroedinger equation ,

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In classical mechanics F = ma is a fundamental law , similarly in quantum mechanics Schroedinger's equation is fundamental equation.

We find of develop Schroedinger equation on the basis of observation and concepts.

From uncertainty principal ―

∆x∆px ≈ ℏ/2

∆x ≈ ℏ/2∆px

If ∆px →0 , then ∆x →∞

Means , uncertainty in momentum is zero then uncertainty in position is infinite , so if momentum of particles is exactly measured then uncertainty in position is infinite

So , particles can go from - ∞ to + ∞ such particle is known as free particle.

So ,

∆x∆(ℏk) ≈ ℏ/2

∆k ≈ 1/2∆x

For free particle , ∆x →∞

So , ∆k = 0

We know that

k = 2π/λ

∆k = -2π∆λ/λ2

So , ∆λ ≈ 0

λ ≈ constant

λ1 ≈ λ2

So , it represent a wave which is coming from - ∞ and going to + ∞ with constant length. Such wave is known as plane wave.

So , wave function corresponding to free particle is plane wave.

So , there four possibility to plane wave solution.

Ψ ( x , t ) ∝ i ei( kx - wt )

Ψ ( x , t ) ∝ i e-i( kx - wt )

Ψ ( x , t ) ∝ Sin ( kx - wt )

Ψ ( x , t ) ∝ Cos ( kx - wt )

So , Now we develop a differential equation because it is easy to solve , this differential equation should follow below condition :

● This differential equation should be linear because quantum state is a linear vector in linear vector space.

● wave is coming from - ∞ to + ∞ so it possibility cab be linear combination above possibility.

● Differential coffcient of linear differential equation should not be function w & k , because we have taken concept of wave packet.

In starting physicists considered wave equation ,

∂2Ψ( x ,t )/∂2x = 1/𝛾2[ 2∂2Ψ( x ,t )/∂2t ] ――( 1 )

Let , Ψ ( x , t ) = Ai ei( kx - wt )

∂Ψ/∂x = ikAei(wt -kx )

∂2Ψ/∂x2 = ( ik )2Ψ = - k2Ψ

∂Ψ/∂t = - iwAei( wt - kx )

∂2Ψ/∂x2 = ( - iw )2Ψ = - w2Ψ

Put above values in equation ( 1 ) .---

-k2Ψ( x , t ) = 1/𝛾2[ ( -w )2Ψ( x , t )

k2 = w2/𝛾2

𝛾 = w/k

Vp = w/k

So , differential coefficient 𝛾 is function of w & k so it represent phase velocity .

But , we have take concept if wave packet so. Wave equation is not valid in quantum mechanics.

However, EM-wave satisfy that equation so, we can say that these wave associate with matter particles are not EM-wave , these are known as matter waves

So further , Schroedinger's considered form of differential equation .

∂2Ψ( x ,t )/∂2x = 1/𝛾2[ 2∂Ψ( x ,t )/∂t ] ―――( 2 )

Now. These are only two possibility of wave function for free particle.

Ψ ( x , t ) ∝ i ei( kx - wt )

Ψ ( x , t ) ∝ i e-i( kx - wt ) and other possibilities

Ψ ( x , t ) ∝ Sin ( kx - wt )

Ψ ( x , t ) ∝ Cos ( kx - wt ) Are not valid because they represent single wave .

So , possibilities

Ψ ( x , t ) ∝ Sin ( kx - wt )

Ψ ( x , t ) ∝ Cos ( kx - wt )

Are superposition principal of sine and cosine so they , represent group velocity

Let , Ψ ( x , t ) = Ai ei( kx - wt )

∂Ψ/∂t = ( - iw )Ψ

∂2Ψ( x , t )/∂x2 = ( ik )2Ψ = - k2Ψ( x , t )

Put above values in equation ( 2 ) .

-k2Ψ ( x , t ) = - iw/𝛾2 [ Ψ ( x , t ) ]

k2 = iw/𝛾2

𝛾2 = iℏ(ℏw)/ℏ2k2 = iℏE/P2 = iℏ/P2[P2/2m ]

𝛾2 = iℏ/2m

So , it is independent of w & k so , it is valid in quantum mechanics.

∂2Ψ( x ,t )/∂2x = 1/𝛾2[ 2∂Ψ( x ,t )/∂t ]

Put value of 𝛾 in above equation ( 2 ).

iℏ/2m[ ∂2Ψ( x ,t )/∂2x ] = ∂Ψ( x , t )/∂t

-ℏ2/2m [ ∂2Ψ( x ,t )/∂2x ] = iℏ [ ∂Ψ( x , t )/∂t ]

Is is Schroedinger's. equation for free particle in one dimension .

In 3-dimension ,

-ℏ2/2m [ ∇ Ψ ( r , t ) = iℏ [ ∂Ψ( r , t )/∂t ]

For free particle ――

-ℏ2/2m [ ∂2Ψ( x ,t )/∂2x ] = iℏ [ ∂Ψ( x , t )/∂t ]

Px2/2m [ Ψ( x ,t ) ] = iℏ [ ∂Ψ( x , t )/∂t ]

Hamiltonian for free particle.

H = k.E. + P. E. , H = Px2/2m

Then HΨ ( x , t ) = iℏ [ ∂Ψ( x , t )/∂t ]

If particle are not free ――

F = ∇V = - ∂V(x)/∂x , then Hamiltonian

H = Px2/2m + V(x)

So , Px2/2m + V(x) [ Ψ ( x , t ) ] = iℏ [ ∂Ψ( x , t )/∂t ]

-ℏ2/2m [ ∂2Ψ( x ,t )/∂2x ] + V(x)Ψ ( x , t ) = iℏ [ ∂Ψ( x , t )/∂t ]

-ℏ2/2m [ ∇2Ψ( r ,t ) + v( r ) Ψ( r ,t ) = iℏ [ ∂Ψ( x , t )/∂t

It is time depends Schroedinger's equation in 3 - D .

NOTE :: 𝛾2 = iℏ/2m , differential coefficient depended Schroedinger's on mass but , in relativistic quantum mechanics mass vairy.

m = m0/√(1 − v2/c2)

So Schroedinger's equation not valid in relativistic quantum mechanics .

● Properties of Schroedinger equation ::

●Schroedinger equation is a linear differential equation in position and time

●Wave function Ψ ( x , t ) should be continuous

●First Derivatives of wave function should be continuous

●Schroedinger equation can describe motion of a single particle . It is not valid for many body system. Howeve, we can use it for two body problem also by converting two body problem into one body using reduce mass.

●Schroedinger equation not valid in relativistic quantum mechanics.

●If Ψ1 ( x , t ) and Ψ2 ( x , t ) are solution of Schroedinger equation then their Linear combination

Ψ ( x , t ) = C1Ψ 1( x , t ) + C2Ψ2 ( x , t ) , also solution of Schroedinger equation.