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Oh sorry A is the amplitude of the ground state wavefunction (I put A in my working to simplify the maths - didn't work)

[tex]A = \frac{1}{\sqrt{\pi a^3}}[/tex]

[tex]A = \frac{1}{\sqrt{\pi a^3}}[/tex]

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[tex]A = \frac{1}{\sqrt{\pi a^3}}[/tex]

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Anyone? I REALLY need this. Please?

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Physics Monkey

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Also, your integration bounds are messed up, you have them flipped for some reason. I didn't check all your constants so I can't say if you have those right.

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[tex] -\int^0_\pi {exp(\frac{prz}{\hbar})sin\theta.dz}[/tex]

I think (I've not tried it on paper just my head so far).

However. Even though the new variable of integration is z, can I still treat the [tex]\theta[/tex] as a constant as this was this orginal variable of interest?

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Sorry my brain burped. Integratiuon by substitution is probably my weakest point in calculus.

Anyway, let me show you how it is as I get it, and maybe you can let me know where I'm going wrong.

The tricky theta integral is

[tex]\int^\pi_0 {e^(\frac{-ipCos\theta}{\hbar})Sin\theta}.d\theta[/tex]

If we make the substitution [tex] z = Cos\theta [/tex]

Then [tex] \frac{dz}{d\theta} = -Sin\theta [/tex]

Now as I understand the integration by substitution process, we need to sub [tex]d\theta[/tex] in terms of dz i.e

[tex] d\theta = \frac{-dz}{Sin\theta} [/tex]

This makes our integral.

[tex]\int^\pi_0 {e^(\frac{-ip.z}{\hbar})Sin\theta}.\frac{dz}{Sin\theta}[/tex]

Now would the Sine terms cancel leaving?

[tex] \int^\pi_0 {e^(\frac{-ip.z}{\hbar})dz [/tex]

Which I can integrate, but not 100% sure about the method? Any bugs in the system?

Anyway, let me show you how it is as I get it, and maybe you can let me know where I'm going wrong.

The tricky theta integral is

[tex]\int^\pi_0 {e^(\frac{-ipCos\theta}{\hbar})Sin\theta}.d\theta[/tex]

If we make the substitution [tex] z = Cos\theta [/tex]

Then [tex] \frac{dz}{d\theta} = -Sin\theta [/tex]

Now as I understand the integration by substitution process, we need to sub [tex]d\theta[/tex] in terms of dz i.e

[tex] d\theta = \frac{-dz}{Sin\theta} [/tex]

This makes our integral.

[tex]\int^\pi_0 {e^(\frac{-ip.z}{\hbar})Sin\theta}.\frac{dz}{Sin\theta}[/tex]

Now would the Sine terms cancel leaving?

[tex] \int^\pi_0 {e^(\frac{-ip.z}{\hbar})dz [/tex]

Which I can integrate, but not 100% sure about the method? Any bugs in the system?

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Physics Monkey

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[tex]-\int^{-1}_{1} {e^(\frac{-ip.z}{\hbar})}.dz[/tex]

Right?

However, I was wondering, if I had made the same substitution but the Sine terms had not cancelled (so ther was still a sine theta in there). Could I have proceeded because I'd now have a a substituted variable and the old variable that need integrating? Could I have treated the theta terms as constant despite theta being the variable of interest?

Right?

However, I was wondering, if I had made the same substitution but the Sine terms had not cancelled (so ther was still a sine theta in there). Could I have proceeded because I'd now have a a substituted variable and the old variable that need integrating? Could I have treated the theta terms as constant despite theta being the variable of interest?

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Right so I solved the theta integral, leaving my r integral as (subbing k for p/h and alpha for 1/a):

[tex] \phi(p) = \frac{2A}{(2\pi/\hbar)^{3/2}}\frac{2\pi}{k}\int^\infty_0{rSin(kr)e^{-\alpha r}}.dr[/tex]

I tried to solve the integral by parts by subbing u =r and using the standard integrals.

[tex]\int{e^{ax} Sin(bx).dx = \frac{e^{ax}}{a^2+b^2}(aSin(bx)-bCos(bx)[/tex]

and

[tex]\int{e^{ax} Cos(bx).dx = \frac{e^{ax}}{a^2+b^2}(bSin(bx)+aCos(bx)[/tex]

Which I'm not sure about. I get (after fiddling with the constants)

[tex] \frac{2\sqrt{2}}{\pi k(\hbar^3a^3)^{1/2}}[\frac{k\alpha}{(k^2-\alpha^2)^2}] [/tex]

Does this sound about right. Can anyone suggest a way to simplify this more?

Whats the forum stand on checking working? I tried the earlier expectation for the kinetic energy of the hydrogen ground state problem and am very close, except I'm a factor of 2 out and can't see where I lost it.

[tex] \phi(p) = \frac{2A}{(2\pi/\hbar)^{3/2}}\frac{2\pi}{k}\int^\infty_0{rSin(kr)e^{-\alpha r}}.dr[/tex]

I tried to solve the integral by parts by subbing u =r and using the standard integrals.

[tex]\int{e^{ax} Sin(bx).dx = \frac{e^{ax}}{a^2+b^2}(aSin(bx)-bCos(bx)[/tex]

and

[tex]\int{e^{ax} Cos(bx).dx = \frac{e^{ax}}{a^2+b^2}(bSin(bx)+aCos(bx)[/tex]

Which I'm not sure about. I get (after fiddling with the constants)

[tex] \frac{2\sqrt{2}}{\pi k(\hbar^3a^3)^{1/2}}[\frac{k\alpha}{(k^2-\alpha^2)^2}] [/tex]

Does this sound about right. Can anyone suggest a way to simplify this more?

Whats the forum stand on checking working? I tried the earlier expectation for the kinetic energy of the hydrogen ground state problem and am very close, except I'm a factor of 2 out and can't see where I lost it.

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- #37

Physics Monkey

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Right, well the k in the constant and the numerator cancel. So my problem must be the extra alpha. I'll take a look at the working again. However can you tell me if there is anything wrong with using that standard integral to do the transform in r?

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Nope still have that pesky alpha, and I feel like I'm so close too.

Can anyone see where I'm going wrong?

Starting with

[tex] \phi(p) = \frac{2A}{(2\pi/\hbar)^{3/2}}\frac{2\pi}{k}\int^\infty_0{rSin(kr)e^{-\alpha r}}.dr[/tex]

Concentrating on the integral, I used by parts making the allocating the following.

[tex] \frac{dv}{dr} = e^{-\alpha r}Sin(kr)[/tex] thus using the standard integral I mentioned I get [tex] v = \frac{e^{-\alpha r}}{k^2-\alpha^2}(- \alpha Sin(kr)-kCos(kr))[/tex]

[tex]u = r [/tex] thus [tex] \frac{du}{dr} = 1 [/tex]

Subbing this into the by parts formula I get.

[tex] [\frac{re^{-\alpha r}}{k^2-\alpha^2}(-\alpha Sin(kr) - kCos(kr))] - \frac{- \alpha}{k^2- \alpha^2}\int{e^{-\alpha r}Sin(kr)}.dr -\frac{k}{k^2-\alpha^2}\int{e^{-\alpha r}Cos(kr)}.dr[/tex]

Making the integration gives me three terms, again, using the standard integrals.

[tex][\frac{re^{-\alpha r}}{k^2-\alpha^2}(-\alpha Sin(kr) - kCos(kr))] - [\frac{- \alpha}{k^2-\alpha^2}}(-\alpha Sin(kr) - kCos(kr))] - [\frac{k}{k^2 - \alpha^2}(-\alpha Cos(kr) + kSin(kr))][/tex]

The terms have limits between zero and infinity. The infinity terms cancel as exp(-infinity) is zero. So only the zero substitutions count. As r = 0 the first term is cancelled to zero. And as Sin zero is zero only the cos terms have anny significance. The Cos reduced to 1 and so I'm left with.

[tex]\frac{- \alpha}{k^2- \alpha^2}(\frac{k}{k^2-\alpha^2}) - \frac{k}{k^2-\alpha^2}({\frac{\alpha}{k^2 -\alpha^2})[/tex]

Leaving me with [tex]\frac{2k \alpha}{(k^2-\alpha^2)^2} [/tex]

And thus the same answer as before. The 2 belongs and the k cancels, its just the alpha that shouldn't be there that I can see. I can't see my error (I'm terrible at finding mistakes) can any of you guys (Assuming what I've written makes sense).

Thanks

Can anyone see where I'm going wrong?

Starting with

[tex] \phi(p) = \frac{2A}{(2\pi/\hbar)^{3/2}}\frac{2\pi}{k}\int^\infty_0{rSin(kr)e^{-\alpha r}}.dr[/tex]

Concentrating on the integral, I used by parts making the allocating the following.

[tex] \frac{dv}{dr} = e^{-\alpha r}Sin(kr)[/tex] thus using the standard integral I mentioned I get [tex] v = \frac{e^{-\alpha r}}{k^2-\alpha^2}(- \alpha Sin(kr)-kCos(kr))[/tex]

[tex]u = r [/tex] thus [tex] \frac{du}{dr} = 1 [/tex]

Subbing this into the by parts formula I get.

[tex] [\frac{re^{-\alpha r}}{k^2-\alpha^2}(-\alpha Sin(kr) - kCos(kr))] - \frac{- \alpha}{k^2- \alpha^2}\int{e^{-\alpha r}Sin(kr)}.dr -\frac{k}{k^2-\alpha^2}\int{e^{-\alpha r}Cos(kr)}.dr[/tex]

Making the integration gives me three terms, again, using the standard integrals.

[tex][\frac{re^{-\alpha r}}{k^2-\alpha^2}(-\alpha Sin(kr) - kCos(kr))] - [\frac{- \alpha}{k^2-\alpha^2}}(-\alpha Sin(kr) - kCos(kr))] - [\frac{k}{k^2 - \alpha^2}(-\alpha Cos(kr) + kSin(kr))][/tex]

The terms have limits between zero and infinity. The infinity terms cancel as exp(-infinity) is zero. So only the zero substitutions count. As r = 0 the first term is cancelled to zero. And as Sin zero is zero only the cos terms have anny significance. The Cos reduced to 1 and so I'm left with.

[tex]\frac{- \alpha}{k^2- \alpha^2}(\frac{k}{k^2-\alpha^2}) - \frac{k}{k^2-\alpha^2}({\frac{\alpha}{k^2 -\alpha^2})[/tex]

Leaving me with [tex]\frac{2k \alpha}{(k^2-\alpha^2)^2} [/tex]

And thus the same answer as before. The 2 belongs and the k cancels, its just the alpha that shouldn't be there that I can see. I can't see my error (I'm terrible at finding mistakes) can any of you guys (Assuming what I've written makes sense).

Thanks

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Anyone see my error, this needs to be in tommorrow

Thanks

Thanks

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- #42

Physics Monkey

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You might try to evaluate the integral by breaking the [tex] \sin{kr}[/tex] up into its exponential parts. You should have two integrals to do, one involves [tex] e^{ikr-r/a_0} [/tex] and the other involves [tex] e^{-ik-r/a_0} [/tex] with an important relative minus sign. Each of these integrals is easy to do as it is a standard exponential integration. Once you've done each integral (be very careful with your minus signs), find a common denominator. The form I gave should pop right out.

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if you are doing prof alexanderovs questions for question 1b i get 0.65 (2dp)but im unsure if that is correct everything seemed to fold out nicely but not sure

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you made a error integrating, i do it by my self and the result is:

[tex]

\phi(p) = \frac{4\pi}{(2\pi\hbar)^{3/2}}\frac{1}{\pi a^3}\frac{2a^3 \hbar ^4}{\left(\hbar ^2 + a^2 p^2 \right)^2}

[/tex]

[tex]

\phi(k) = \frac{4\pi}{(2\pi\hbar)^{3/2}}\frac{1}{\pi a^3}\frac{2a^3 \hbar ^4}{\left(\hbar ^2 +\hbar ^2 a^2 k^2 \right)^2}

[/tex]

[tex]

\phi(p) = \frac{4\pi}{(2\pi\hbar)^{3/2}}\frac{1}{\pi a^3}\frac{2a^3 \hbar ^4}{\left(\hbar ^2 + a^2 p^2 \right)^2}

[/tex]

[tex]

\phi(k) = \frac{4\pi}{(2\pi\hbar)^{3/2}}\frac{1}{\pi a^3}\frac{2a^3 \hbar ^4}{\left(\hbar ^2 +\hbar ^2 a^2 k^2 \right)^2}

[/tex]

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