Wavefunction of a Hydrogen atom is expressed in eigenfunctions as: $$\psi(\boldsymbol r,t=0)=1/\sqrt{14}(2\psi_{100}(\boldsymbol r)-3\psi_{200}(\boldsymbol r)+\psi_{322}(\boldsymbol r) ).$$

  1. Is this an eigenfunction of parity operator?

  2. What is probability of finding system in state (100),(200),(322)?

  3. In another energy eigenstate?

  4. What is the expectation value of energy, $L^2$,$L_z$?

I know some algebra but I don't know what the subscript of eigenfunction means ._. The only thing I know is about parity operator and its eigenfunctions. Others are just like foreign language.


closed as off-topic by jinawee, Qmechanic Apr 7 '14 at 23:07

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Just some things which might help:

  • the subscript of $\psi_{nlm}$ refer to the principal quantum number $n$, the azimuthal quantum number $l$, and the magnetic quantum number $m$; the properties of last two you can understand from properties of the spherical hamronics $Y_{lm}$ by noting that $\psi_{nlm}(r,\theta,\phi) = R_{nl}(r) Y_{lm}(\theta,\phi)$.

  • the parity operator $P$ acts like $P: \mathbf{r}\mapsto -\mathbf{r}$ which corresponds to $\theta\mapsto\pi-\theta$ and $\phi\mapsto \pi+\phi$. It is easy to convince yourself that $Y_{lm}(\pi-\theta,\phi) = (-1)^l Y_{lm}(\theta,\phi)$ such that $P \psi_{nlm} = (-1)^l \psi_{nlm}$.

  • also from the properties of the spherical harmonics, you can show that $$\langle \psi_{nlm} | L^2 | \psi_{nlm} \rangle = \hbar^2 l (l+1)$$ and $$\langle \psi_{nlm} | L_z | \psi_{nlm} \rangle = \hbar m.$$

  • the answers to your questions follow easily from the statements above, the linearity of expectation values, and from the fact that states $\psi_{nlm}$ with different $(nlm)$ are orthogonal to each other.

  • $\begingroup$ THank you, just a small question: so it s ok to use $\langle\psi_n|E|\psi_n\rangle=-13.6eV/n^2$ ? $\endgroup$ – Geralt of Rivia Nov 13 '12 at 15:27
  • $\begingroup$ @Lorenz: yes, the energy depends only on $n$ in the fashion you have indicated. $\endgroup$ – Fabian Nov 13 '12 at 19:36

To answer this question, you don't actually have to know the meaning of those subscripts. First I'll try and point you in the right direction -

To find the probability of say, $(100)$ occurring, you can think of a (hopefully) more familiar example of a state (say $|\psi\rangle$) given by $$|\psi\rangle = \frac{|a\rangle}{\sqrt{2}} + \frac{|b\rangle}{2} + \frac{|c\rangle}{2}$$

Looking at it, I'm sure you'll be able to guess the probability that when $|\psi\rangle$ is observed, it falls into either $|a\rangle$, $|b\rangle$, or $|c\rangle$. It's an exactly analogous situation in your problem. Once you've calculated the probability of each energy eigenstate occurring, it's trivial to calculate the probability of it being in any other state except those three.


Coming to what the numbers actually mean - If you see the eigenstate of a hydrogen atom written as $\psi_{100}(\vec r)$, the $(100)$ means that the electron in question has the quantum numbers $n = 1$, $l = 0$, $m = 0$.

Now, the quantum number $n = 1$ means that the electron is in the first orbital (also referred to as the $s$ orbital), $l = 0$ means the orbital angular momentum eigenvalue is zero, and $m = 0$ means the spin angular momentum eigenvalue is zero. The $n$, $l$ and $m$ values come from a solution of the schrodinger equation with a spherically symmetric coloumb potential. If you know what spherical harmonics are, you can see that the $l$ and $m$ values are the same as the $Y_{l}^{m}(\theta, \phi)$.


You need to calculate the expectation of $E$, $L_{z}$ and $L^{2}$. As you probably know, the equations for that are given by $$\langle E \rangle = \langle \psi| E |\psi \rangle$$ $$\langle L^{2} \rangle = \langle \psi | L^{2} |\psi \rangle $$ $$\langle L_{z} \rangle = \langle \psi | L_{z} | \psi \rangle $$

So clearly all you need to know is how each one of those operators acts on $|\psi \rangle$ and you'll get your answer.

  • $\begingroup$ P(100)=2/7, P(200)=9/14, P(322)=1/14 P(otherstate)=0 i already knew this. How to check if this is an eigen function of parity ( if it's an even or odd function) ? i feel i need to understand each subscript and how they behave under parity operator $\endgroup$ – Geralt of Rivia Nov 13 '12 at 13:03
  • $\begingroup$ So what was your question? $\endgroup$ – Kitchi Nov 13 '12 at 13:05
  • $\begingroup$ also, is there any short, elegant way to calculate $\Psi*E\Psi$, $\Psi*L^2\Psi$, $\Psi*L_z\Psi$, Thank you ~ $\endgroup$ – Geralt of Rivia Nov 13 '12 at 13:10

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