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A hydrogen atom is characterized by the wavefunction

$$\mid \psi \rangle =\sqrt{\frac{2}{7}}\mid 4\,2\,1\rangle +\sqrt{\frac{1}{7}}\mid 2 \,1\,\bar{1}\rangle+\sqrt{\frac{4}{7}}\mid 3\,2\,0\rangle$$

I want to know the possible outcomes and the probability for obtaining each outcome from measuring $E$, $L^2$, and $L_z$, while also calculating the expectation value of the energy expressed as some number times $E_1$ (for $E$), $\hbar^2$ (for $L^2$), and $\hbar$ (for $L_z$). I know the possible outcomes are the eigenvalues for the observable, but we didn't really go over any examples, and there are few examples of this sort in the textbook, so I don't know where to begin. For example, I know when we act $L^2$ on the system, we will get

$$ L^2\mid \psi\rangle=6\hbar^2\sqrt{\frac{2}{7}}\mid 4\,2\,1\rangle +2\hbar^2\sqrt{\frac{1}{7}}\mid 2 \,1\,\bar{1}\rangle+6\hbar^2\sqrt{\frac{4}{7}}\mid 3\,2\,0\rangle$$

However, what does this tell me about the possible outcomes? Would the probabilities for each outcome be the coefficient multiplying each successive term squared divided by the Pythagorean sum of the all three terms? How do I find the expectation value? Any help would be appreciated.

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By the way, what you have there is not a wave function, it is a state ket. The wave function is just the coefficient of the position basis. –  silvrfück Oct 27 '13 at 20:28
    
According to the problem, $\mid \psi\rangle$ listed above is a wave function. This is getting very confusing--do you know of any references (besides Griffiths) that goes into senior-level quantum mechanics more in depth? –  Bronzeclocksofbenin Oct 27 '13 at 20:43
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Well, sometimes the term wave function is used in a rather careless way. A quantum state is characterized by a vector. If a basis is given, a vector can be characterized by its coordinates in that basis, and the wave function is nothing but the function that gives the coordinates of your state ket in the position basis. So the same information that it is in the ket is also in the wave function, that is why the term is used ,at least in my opinion, carelessly. For quantum mechanics i strongly reccomend Cohen-Tanoudji's book, in my opinion, it is really well written. –  silvrfück Oct 28 '13 at 11:17
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The state ket of a system has encoded all the information that can be known about that system. In which way? Well, in its linear decomposition on the eigenstates that form a basis in the Hilbert space you are working at.

Your particular ket is represented as a linear combination of 4 kets who have nonzero coefficients. All of those kets represent a particular basis-state(which is characterized using complete sets of commuting obsevables, in your case $E,L^2$ and $Lz$). The 3 numbers "inside" the ket give a possible values of $E,L^2$ and $Lz$(via eigenvalue equations). With what probability?well that is just the modulus of the coefficient in the linear composition. So you see, the possible values are the ones without zero coefficient, that is, the ones with nonzero probability.

Now, if you wanted information about another magnitude apart from $E,L^2$ or $Lz$, you would have to express your current base kets in the eigenkets(in another complete set of commuting observables) of the magnitude whose information you wanna know.

Beware normalization always(your ket is normalized)

To get the expectation value you just "Bra Ket" your operator(if you think carefully you will realize that by this you just make a sum of all the possible values multiplied by their probabilities).

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How do the numbers "inside" the ket (namely, $n$, $\ell$, and $m_l$) give me the possible values of $E$, $L^2$, and $L_z$? For example, I know $L^2\mid n\ell m\rangle = \ell(\ell+1)\hbar^2 \mid n\ell m\rangle$; does not mean the possible values of measuring $L^2$ are simply $6\hbar^2$, $2\hbar^2$, and $6\hbar^2$? –  Bronzeclocksofbenin Oct 27 '13 at 19:54
    
yeah, exactly. Just use the eigenvalue equation. –  silvrfück Oct 27 '13 at 20:22
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