Fourier transform of the ground state hydrogen wave function

[Ado]

Hi!

1. Homework Statement
From the website http://www1.uprh.edu/rbaretti/MomentumspaceIntegration8feb2010.htm
we can see the Fourier transform of the ground state hydrogenic wave function :

Φ(p) = ∫ ∫ ∫ exp(-i p r) (Z³/π )^1/2 exp(-Zr) sin(θ) dθ dφ r² dr (1.1)

After intregation of the variable φ we have :

Φ(p) = 2π (Z³/π )^1/2 ∫ ∫ { exp(-i pr cos(θ ) sin(θ) dθ } exp(-Zr) r² dr (1.2)

and next :

Φ(p) = 2π (Z³/π )^1/2 ∫ { 2 sin(pr) /pr } exp(-Zr) r² dr (1.3)

= 4π (Z³/π )^1/2 ∫ {sin(pr) /p} exp(-Zr) r dr

= 8 π^1/2/ Z^5/2 /( p² + Z² )²

And I want to understand these different steps..

Homework Equations

I don't understand the relations used between (1.1), (1.2) and (1.3) :
(1.1) to (1.2) new cos(θ ) appears
(1.2) to (1.3) sin(pr)/pr appears

The Attempt at a Solution

Can you explain me what happen between (1.1) and (1.3) ?

Tanks in advance !

 

[mjc123]

In 1.1 we have exp(-ip.r). p.r is the dot product of two vectors, whose value is pr*cosθ, where θ is the angle between the vectors. I suppose in p space, which I'm not familiar with, this angle must be equal to the coordinate θ. There is a bracket missing in 1.2, which should read
Φ(p) = 2π (Z³/π )^1/2 ∫ ∫ { exp(-i pr cos(θ) ) sin(θ) dθ } exp(-Zr) r² dr (1.2)
This can be integrated over θ by using the substitution u = cosθ and the identity e^ix = cosx + i*sinx, to give 1.3

 

[Ado]

Thanks for your reply mjc123 !

I had applied your recommendation and I find the result in (1.3)

If I used the substitution u = cosθ, I must then integrate between 1 and -1 (cos(0) = 1 and cos(pi) = -1).

I put du = -sinθdθ, so :

exp(-i pr cos(θ) ) sin(θ) dθ = -exp(-i pr u) du = - cos(-pru)du + isin(pru)du

The complex term disappears with the integration and

∫₁^-1 - cos(-pru) du = 2sin(pr)/pr

I have just a question about the bounds of integration. At the beginning, we integrate between 0 and pi and with the substitution, between 1 and -1. The order of these bounds is important because if we integrate du between -1 and 1 we would have a negative term. My question is probably stupid but why are we constrained at the beginning to integrate between 0 and pi and not between pi and 0 ??

Thanks in advance !

 

[mjc123]

A mathematician could no doubt give you a better formal explanation, but I would say basically: because the increment is dθ. You could also integrate from pi to 0 with increment -dθ.
If you think of it in terms of the area under a curve, for example, suppose you have a function y which is positive over the range x = 0 to 1, and you want the area under this curve, which of course is positive. This is ∫₀¹ydx. We could also go from x = 1 to 0 in increments of -dx, and get ∫₁⁰y(-dx) = ∫₁⁰(-y)dx = ∫₀¹ydx. Or think of the function y as the differential of the integral, i.e. the rate of change of the area with x. This is obviously positive (or negative if y is negative) with increasing x, i.e. positive dx. So we integrate y by dx over the interval 0 to 1, not 1 to 0. However, if we do a coordinate transformation, we must keep the initial and final limits in their transformed form, and transform dx correctly, so if we made the substitution u = e^-x, we would integrate -y(u)/u*du from 1 to e^-1.
Hope this is not too rubbish.

 

[Ado]

Thanks a lot for your explication mjc123 and thanks for your help!

 

[loued]

Hi!

1. Homework Statement
From the website http://www1.uprh.edu/rbaretti/MomentumspaceIntegration8feb2010.htm
we can see the Fourier transform of the ground state hydrogenic wave function :

Φ(p) = ∫ ∫ ∫ exp(-i p r) (Z³/π )^1/2 exp(-Zr) sin(θ) dθ dφ r² dr (1.1)

After intregation of the variable φ we have :

Φ(p) = 2π (Z³/π )^1/2 ∫ ∫ { exp(-i pr cos(θ ) sin(θ) dθ } exp(-Zr) r² dr (1.2)

and next :

Φ(p) = 2π (Z³/π )^1/2 ∫ { 2 sin(pr) /pr } exp(-Zr) r² dr (1.3)

= 4π (Z³/π )^1/2 ∫ {sin(pr) /p} exp(-Zr) r dr

= 8 π^1/2/ Z^5/2 /( p² + Z² )²

And I want to understand these different steps..

Homework Equations

I don't understand the relations used between (1.1), (1.2) and (1.3) :
(1.1) to (1.2) new cos(θ ) appears
(1.2) to (1.3) sin(pr)/pr appears

The Attempt at a Solution

Can you explain me what happen between (1.1) and (1.3) ?

Tanks in advance !

How was the Sin integral calculated in the last step?

 

[PeroK]

How was the Sin integral calculated in the last step?

Note that this homework is from over four years ago. Hopefully Ado has graduated by now.

 

[Vanadium 50]

Or at least turned it in!

 

[mjc123]

How was the Sin integral calculated in the last step?

Use the fact that sin(pr) is the imaginary part of e^ipr.

 

[vela]

Or integrate by parts.