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Suppose you have a mix of states made up of the Hydrogen $\lvert nlm \rangle$ states where one of the coefficients is unknown. For example: $$ \lvert \psi\rangle=A\lvert 100\rangle + \sqrt{\frac{2}{3}}|210> + \sqrt{\frac{2}{3}}\lvert 211\rangle - \sqrt{\frac{2}{3}}\lvert 21-1\rangle $$ Since all the hydrogen wavefunctions $\lvert nlm \rangle$ are orthonormal I suppose that the normalization condition does not work for the above example since $$ \lvert A\rvert^2+\frac{2}{3}+\frac{2}{3}+\frac{2}{3} =\lvert A \rvert^2+2 > 1 $$ regardless the value of $A$. Even if one takes $A=i$ (purely imaginary) still the normalization does not work.

  • Am I missing something here? Can the above function be normalized by a particular choice of $A$ coefficient?
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  • $\begingroup$ Please consider using MathJax for the equations and symbols in the future. $\endgroup$ – Phonon Apr 23 '15 at 14:02
  • $\begingroup$ So I can use directly Latex here at Stackexchange when writing the equations? I did not know. Thank you for the tip. $\endgroup$ – user36636 Apr 23 '15 at 14:30
  • $\begingroup$ I recently learned you can even define \newcommand{\ket}[1]{| #1 \rangle}, which is helpful for typing up QM stuff. $\endgroup$ – zeldredge Apr 23 '15 at 14:36
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The answer is that the premise is wrong. There can't be a hydrogen wave function with the coefficients you have written. Even if there was no $| 1 0 0 \rangle$ state present, the state isn't normalized. That means that it isn't physical. However, remember that the coefficients are somewhat arbitrary, that is, we're allowed to multiply the whole wavefunction by some constant $C$. That's how we normalize them in the first place. So the important thing about your state isn't the $\sqrt{\frac{2}{3}}$ part, it's the fact that all the other coefficients are the same, have equal probability. So you could write something like $$ \newcommand{\ket}[1]{| #1 \rangle} \ket{\psi} =A\ket{100} + B\ket{210} + B\ket{211} - B\ket{21-1} $$

$$ |A|^2+3 |B|^2 =1 \\ |A|^2 = 1 - 3 |B|^2 $$ So from this you can see the condition on $B$ for your example to make sense, we need $|B|^2 < 1/3$. (We can get a negative square from imaginary numbers, but never a negative absolute square.) Basically, there needs to be some probability left over for the other state you want to insert. Give me that probability for the other three states and I can tell you the magnitude of the other one, but I can't do that for the state you provided.

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  • $\begingroup$ Yes this is also what I supposed on the first place that there must be some constant that multiply all the initial wavefunction and make it "normalizable". Especially because the text of this, requires further the probabilities of energies that can be measured. But the given function lacks this "overall" multiplication constant for all terms and one is required on the first place to find the value of A that normalizes the initial wavefunction. $\endgroup$ – user36636 Apr 23 '15 at 14:26
  • $\begingroup$ Wait, this is a textbook problem? That's very odd then, since it seems poorly posed enough to be un-answerable in the form you gave. $\endgroup$ – zeldredge Apr 23 '15 at 14:31
  • $\begingroup$ I'm not a student, I'm a tutor (pretty good at QM) and one of my students (from NY) game me his HW. This was the last question and I just supposed that his professor should be wrong when asking this question, but I needed a second opinion. This is not the first "strange" question that this professor has asked his students. $\endgroup$ – user36636 Apr 23 '15 at 14:34
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The state you have given is not normalisable as a consequence of the results of the calculations you have done. Even if the first state (with coefficient $A$) was not present it would not be normalised. To normalise what you have given, another constant needs to multiply everything through (so that the relative proportions are unchanged

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Looks like textbook hybridization problem, did you check the usual suspects, or e.g. this one?

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