2 added 446 characters in body edited Nov 13 '12 at 13:21 Kitchi 3,16721635 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 $$Y_{l}^{m}(\theta, \phi)$$.   Edit: 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. 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 $$Y_{l}^{m}(\theta, \phi)$$.   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 $$Y_{l}^{m}(\theta, \phi)$$. Edit: 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. 1 answered Nov 13 '12 at 12:07 Kitchi 3,16721635 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 $$Y_{l}^{m}(\theta, \phi)$$.