Superposition of states of the hydrogen atom

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say we had two states [tex]\psi1[/tex] and [tex]\psi2[/tex] and i want to model the superposition of the two states [tex]\psi[/tex]=[tex]c1\psi1[/tex]+[tex]c2\psi2[/tex]. how do i find c1 and c2? ive been trying to do c1=[tex]\int[/tex][tex]\psi[/tex][tex] \psi1[/tex] [tex]r^2dr[/tex] over the limits 0 and infinity but i dont seem to be getting anywhere. does anyone have any ideashow i would do this?

thanks :)
 

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  • #2
George Jones
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Are states [itex]\psi1[/itex] and [itex]\psi2[/itex] normalized? Orthogonal to each other?
 
  • #3
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erm... they are both normailised. and im not sure if they are orthogonal to each other... im doing it for the solutions ofthe hydrogen atom. so they are both solutions of the schrodinger equation for the hyrdogen atom. the only difference between the two wave functions is the principle quantum number, n, and hence the energy.
 
  • #4
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I vaguely recall that the solutions to the hydrogen atom are the Bessel functions. If this is true, then the weight would be r, not r^2, so it would be: [tex]c_1=\int \psi*\psi_1 rdr [/tex].
 
  • #5
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The overall solution of the equation is:

[tex]\Psi[/tex] = [tex](-1)^m\frac{1}{a_0^\frac{3}{2}}\frac{2}{n^2}\left[\frac{(n-\textit{l}-1)!}{(n+\textit{l})!}\right]^{\frac{1}{2}}\left(\frac{2\textbf{r}}{na}\right)^\textit{l}e^{\frac{-\textbf{r}}{na}}L^{2\textit{l}+1}_{n-\textit{l}-1}\left[\frac{2l+1}{4\pi}\frac{(\textit{1}-|m|)!}{(\textit{l}+|m|)!}\right]^{\frac{1}{2}}P^{m}_{l}(cos\theta)[/tex]

where the L is the laguerre polynomial and the P is the associated legendre polynomial.

The two states im trying to super impose is the 1s (n=1, l=0, m=0) and 2s (n=2, l=0, m=0) states. Hence the legendre polynomial will be the same each time, but the other terms will differ slightly.

where does the r in your equation come from? is the '*' the complex conjugate or multiplication and is my approach indeed correct because im not entirely sure where it comes from.
 
  • #6
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I wonder what you are trying to do. The physical pictures for this superposition are pretty easy to work out. The orbital is either compressed or expanded depending on the polarity of the combination. When you throw in the differential time variation of the two states, you get kind of a pulsating balloon for the electron orbital. Because it is spherically symmetric, it doesn't radiate. Which explains why the transition between the two states is forbidden.
 
  • #7
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Since the hydrogen atom states are orthonormal, you just have to calculate the inner product [itex]<\psi_1 | \psi>=\int \psi_1^* \psi d^3 \mathbf r[/itex]. Make sure you include the angular part of the integral!
 
  • #8
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okay...so would that mean that [itex]c1=\int \psi1^*\psi r^2sin \phi dr d\theta d\phi[/itex]? what does this physically mean anyway? also how can i work out this inner product to work out c1 and c2 when c1 and c2 are included in [tex]\psi[/tex] or do i just use [tex]\psi[/tex]=[tex]\psi1+\psi2[/tex] instead of [tex]\psi[/tex]=[tex]c1\psi1+c2\psi2[/tex]? sorry guys...i think this is prob really simple but i just dont get it :(
 
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  • #9
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i meant: [itex]c1=\int \psi1^*\psi r^2sin \phi dr d\theta d\phi[/itex] not [itex]c1=\int \psi1^*\psi r^2sin \theta dr d\theta d\phi[/itex]
 
  • #10
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If you already have [itex]\psi[/itex] explicitly in terms of [itex]c_1[/itex] and [itex]c_2[/itex], why would you need to go through the trouble of performing the integration? Perhaps you could tell us what exactly it is you're trying to do?

(FYI, calculating the inner product is, roughly speaking, finding the amount of overlap that exists between [itex]\psi[/itex] and [itex]\psi_1[/itex].)
 
  • #11
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im trying to model the probability of the superpostition of the two lowest energy states of the hydrogen atom ie 1s and 2s energy levels using mathematica. i have the wavefunction of the 1s energy level and 2s energy level. I no that [itex]\psi = c1\psi1 +c2 \psi2, [/itex] and now no that [itex]c1[/itex]=inner product of [itex]\psi1[/itex]and [itex]\psi[/itex]. I think my main problem is not knowing [itex]\psi[/itex] or c1, c2 so it seems to be a big circle and im getting nowhere. would i use my [itex]\psi[/itex] to be my function at t=0, so eg if the electron is in the 1s state initially, then i should use [tex]\psi[/tex] to be [tex]\psi1[/tex]?
 
  • #12
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im trying to model the probability of the superpostition of the two lowest energy states of the hydrogen atom ie 1s and 2s energy levels using mathematica. i have the wavefunction of the 1s energy level and 2s energy level. I no that [itex]\psi = c1\psi1 +c2 \psi2, [/itex] and now no that [itex]c1[/itex]=inner product of [itex]\psi1[/itex]and [itex]\psi[/itex]. I think my main problem is not knowing [itex]\psi[/itex] or c1, c2 so it seems to be a big circle and im getting nowhere. would i use my [itex]\psi[/itex] to be my function at t=0, so eg if the electron is in the 1s state initially, then i should use [tex]\psi[/tex] to be [tex]\psi1[/tex]?
Why don't you just let c1 = c2 = 1/squrt(2) and see what happens.
 
  • #13
Physically what c1 and c2 tell you (or rather what the squares of c1 and c2 tell you) are the probabilities of finding the particle in the states psi 1 and psi 2 respectively when you make a measurement and it is in the state psi. This should give you another equation involving c1 and c2, namely that c1^2 + c2^2 = 1 (this equation is because the particle must be found in either psi 1 or psi 2 when you make a measurement... the fancy way of saying this is that the particle is thrown into an eigenstate of the observable... and the total probability is 1) Does that help?
 
  • #14
jtbell
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I think my main problem is not knowing [itex]\psi[/itex] or c1, c2 so it seems to be a big circle and im getting nowhere.
Based on the information you've given us so far, all that can be said about [itex]c_1[/itex] and [itex]c_2[/itex] is that if [itex]\psi_1[/itex], [itex]\psi_2[/itex] and [itex]\psi[/itex] are all normalized, then [itex]c_1^2 + c_2^2 = 1[/itex]. Any combination of [itex]c_1[/itex] and [itex]c_2[/itex] that meets this condition is a valid superposition.

Exactly what values of [itex]c_1[/itex] and [itex]c_2[/itex] you should use depends on what you want to model. For example, if you want to model the transition from n = 2 to n = 1, then you could choose [itex]c_1[/itex] and [itex]c_2[/itex] to be functions of t such that initially [itex]c_1 = 0[/itex] and [itex]c_2 = 1[/itex], and finally [itex]c_1 = 1[/itex] and [itex]c_2 = 0[/itex], and [itex]c_1^2(t) + c_2^2(t) = 1[/itex] for all intermediate values of t.

Or if you want a steady-state mixture with equal and constant probabilities for the two states, then you would use Conway's suggestion.
 
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  • #15
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Thanks so much for all your help-ive got a much clearer picture of whats goin on now! however i have a question about the following statement:

For example, if you want to model the transition from n = 2 to n = 1, then you could choose [itex]c_1[/itex] and [itex]c_2[/itex] to be functions of t such that initially [itex]c_1 = 0[/itex] and [itex]c_2 = 1[/itex], and finally [itex]c_1 = 1[/itex] and [itex]c_2 = 0[/itex], and [itex]c_1^2(t) + c_2^2(t) = 1[/itex] for all intermediate values of t.

.
can you have time dependent cooefficients considering the wavewfunction itself statisfies the schodinger equaiton for hydrogen atom:

[itex]\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\left(\psi(x)\phi(t)\right)+V(x)\psi(x)\phi(t)=i\hbar\frac{\partial}{\partial t}\psi(x)\phi(t).[/itex]

and the new superimposed wavefunction also needs to satisfy this equation. wouldnt the wavefunction now have the wrong dimensions etc?

thanks
 
  • #16
jtbell
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The dimensions aren't a problem because [itex]c_1[/itex] and [itex]c_2[/itex] are dimensionless numbers, even if they vary with time.

Making them time dependent does make the superposition fail to satisfy the S.E., strictly speaking. The time derivatives of the coefficients mess things up when you substitute the superposition into the time-dependent S.E. to test whether it's a solution. However, if they change "slowly enough", then I think you can safely assume that the superposition is a "good enough" approximation to an exact solution. We do this kind of "quasistatic approximation" in thermodynamics and electrodynamics, too.
 
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