Having trouble normalizing wave function

[klaustukast]

Electron in hydrogen atom is defined by this wave function :
Ψ(r,ϑ,φ)=Ar²exp(-2r/a)cos²(ϑ)exp(-3iφ)
proton is in the center of the coordinate system.a is a known positive constant.
I'm trying to find normalizing constant A.
Ψ*(r,ϑ,φ)=Ar²exp(-2r/a)cos²(ϑ)exp(3iφ)
I get that ∫∫∫(ψ*)ψdV=1.
dV=r²sinϑ dr dϑ dφ
i get ∫∫∫A²r⁶exp(-4r/a)cos⁴(ϑ)*sin(ϑ) dr dϑ dφ (integrals are definite 0 to π,0 to 2π and 0 to infinity) If I calculate this I should get a constant and get A from that.
Problem I'm having is calculating ∫dr(integral is from zero to infinity) part cause I get ∫(r^6exp(-4r/a))dr which is infinity if I put in infinity.Am I doing something incorrectly or is the given equation Ψ wrong?

 

[Orodruin]

part cause I get ∫(r^6 exp(-4r/a))dr which is infinity if I put in infinity.

What do you mean by "put in infinity"? Please be explicit and show your work.

 

[klaustukast]

What do you mean by "put in infinity"? Please be explicit and show your work.

Integral is definite from zero to infinity.

 

[Orodruin]

Again, please show your work.

 

[klaustukast]

Again, please show your work.

∫(r^6 exp(-4r/a))dr =
a*exp(-4r/a)*(256x⁶ + 384ax⁵ + 480a²x⁴ + 480a³x³ + 360a⁴x² + 180a⁵x+45a⁶)/1024 and then I try to put in 0 and infinity cause its definite integral. And as you can see putting in infinity doesn't give me a constant so i can't find A. It gives me nonsense.

 

[Orodruin]

And as you can see putting in infinity doesn't give me a constant so i can't find A.

The upper bound of the integral is defined as the limit when $r \to \infty$. What happens to your integrated expression in that limit?

(Also, note that you have a overall sign error.)

 

[klaustukast]

The upper bound of the integral is defined as the limit when $r \to \infty$. What happens to your integrated expression in that limit?

(Also, note that you have a overall sign error.)

I posted wrong integral sorry. Correct one =-a*exp(-4r/a)*(256r⁶ + 384ar⁵ + 480a²r⁴ + 480a³r³ + 360a⁴r² + 180a⁵r+45a⁶)/1024
If my r goes to infinity in this case its undefined right?

 

[Ray Vickson]

Electron in hydrogen atom is defined by this wave function :
Ψ(r,ϑ,φ)=Ar²exp(-2r/a)cos²(ϑ)exp(-3iφ)
proton is in the center of the coordinate system.a is a known positive constant.
I'm trying to find normalizing constant A.
Ψ*(r,ϑ,φ)=Ar²exp(-2r/a)cos²(ϑ)exp(3iφ)
I get that ∫∫∫(ψ*)ψdV=1.
dV=r²sinϑ dr dϑ dφ
i get ∫∫∫r⁶exp(-4r/a)cos⁴(ϑ)*sin(ϑ) dr dϑ dφ (integrals are definite 0 to π,0 to 2π and 0 to infinity) If I calculate this I should get a constant and get A from that.
Problem I'm having is calculating ∫dr(integral is from zero to infinity) part cause I get ∫(r^6 =-exp(-4r/a))dra*exp(-4r/a)*(256r6 + 384ar5 + 480a2r4 + 480a3r3 + 360a4r2 + 180a5r+45a6)/1024 which is infinity if I put in infinity.Am I doing something incorrectly or is the given equation Ψ wrong?

There are two different conventions for spherical polar coordinates: (1) $\theta$ is the polar angle (latitude) and $\phi$ equatorial angle (longitude); or (2) $\theta$ is the equatorial angle (longitude) and $\phi$ is the polar angle (latitude). Convention (1) is used in most physics books and papers, while (2) is often used in math books and papers.

Which convention are you employing?

 

[klaustukast]

There are two different conventions for spherical polar coordinates: (1) $\theta$ is the polar angle (latitude) and $\phi$ equatorial angle (longitude); or (2) $\theta$ is the equatorial angle (longitude) and $\phi$ is the polar angle (latitude). Convention (1) is used in most physics books and papers, while (2) is often used in math books and papers.

Which convention are you employing?

Doesnt matter for my problem

 

[Orodruin]

I posted wrong integral sorry. Correct one =-a*exp(-4r/a)*(256r⁶ + 384ar⁵ + 480a²r⁴ + 480a³r³ + 360a⁴r² + 180a⁵r+45a⁶)/1024
If my r goes to infinity in this cases its undefined right?

Not right. What goes to infinity faster, exp(kr) or r^n (for fixed numbers k and n both > 0).

There are two different conventions for spherical polar coordinates: (1) $\theta$ is the polar angle (latitude) and $\phi$ equatorial angle (longitude); or (2) $\theta$ is the equatorial angle (longitude) and $\phi$ is the polar angle (latitude). Convention (1) is used in most physics books and papers, while (2) is often used in math books and papers.

Which convention are you employing?

This is not really relevant for the r-integral, which is what the OP is currently struggling with.

Regardless, the convention is clear from the expressions (physics convention).

 

[klaustukast]

Not right. What goes to infinity faster, exp(kr) or r^n (for fixed numbers k and n both > 0).This is not really relevant for the r-integral, which is what the OP is currently struggling with.

Regardless, the convention is clear from the expressions (physics convention).

Thanks, I've got it.

 

[Orodruin]

Thanks, I've got it.

Would you mind sharing your solution? Others might havd the same question and forum rules prevent discussing the full solution until it is clear that you have solved it.

 

[klaustukast]

Would you mind sharing your solution? Others might havd the same question and forum rules prevent discussing the full solution until it is clear that you have solved it.

Since exp goes to infinity faster I get zero.And from bottom bound: when I put in zero I get 45a⁷/1024

 

[Orodruin]

Right, that is correct.

Now, for completeness, there is a way of computing the integral without finding the explicit primitive function. Namely to do the variable substitution $x = 4r/a$. This leads to the integral
$$
\int_0^\infty r^6 e^{-4r/a}dr = \frac{a^7}{2^{14}} \int_0^\infty x^6 e^{-x}dx =\frac{a^7}{2^{14} }\Gamma(7) =\frac{6! a^7}{2^{14}} = \frac{45a^7}{1024}.
$$
Of course, this requires familiarity with the gamma function, but it is a good idea to get familiar with it as a physicist.

 

[Ray Vickson]

Not right. What goes to infinity faster, exp(kr) or r^n (for fixed numbers k and n both > 0).This is not really relevant for the r-integral, which is what the OP is currently struggling with.

Regardless, the convention is clear from the expressions (physics convention).

Yes, of course, I can guess. But why should I have to?

 

[Orodruin]

Yes, of course, I can guess. But why should I have to?

I don't think it is necessary to guess. It is directly evident from the volume element (or the form of the spherical harmonic if you are so inclined). Besides, it is not really relevant to the OP’s issues presented in this thread - only the radial integral was. So no, I did not have to guess to help the OP with his problem. The answer would have been the same regardless of the convention.