- #1

- 217

- 0

thanks :)

- Thread starter lavster
- Start date

- #1

- 217

- 0

thanks :)

- #2

George Jones

Staff Emeritus

Science Advisor

Gold Member

- 7,394

- 1,030

Are states [itex]\psi1[/itex] and [itex]\psi2[/itex] normalized? Orthogonal to each other?

- #3

- 217

- 0

- #4

- 970

- 3

- #5

- 217

- 0

[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

- 397

- 0

- #7

- 468

- 4

- #8

- 217

- 0

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 :(

Last edited:

- #9

- 217

- 0

- #10

- 468

- 4

(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

- 217

- 0

- #12

- 397

- 0

Why don't you just let c1 = c2 = 1/squrt(2) and see what happens.

- #13

- 111

- 0

- #14

jtbell

Mentor

- 15,664

- 3,733

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.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.

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.

Last edited:

- #15

- 217

- 0

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

.

[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

Mentor

- 15,664

- 3,733

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.

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.

Last edited:

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 20K

- Last Post

- Replies
- 1

- Views
- 9K

- Last Post

- Replies
- 12

- Views
- 444

- Replies
- 1

- Views
- 784

- Last Post

- Replies
- 3

- Views
- 485

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 642

- Replies
- 3

- Views
- 967

- Last Post

- Replies
- 10

- Views
- 771