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Principal quantum number n is an integer, Why?

  1. Jul 16, 2015 #1
    Why is it necessary that the principal quantum number of a Hydrogen atom problem in Quantum mechanics must be an integer?Couldn't it be any fraction?
     
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  3. Jul 16, 2015 #2

    mfb

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    Plugging fractions into the equations does not lead to solutions of the Schroedinger equation. It's like trying to find the 1.6th person in a row: that does not make sense.
     
  4. Jul 16, 2015 #3

    Nugatory

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    Google for "Hydrogen atom Schrodinger's equation" to see why the principal (and others as well) quantum number must be an integer.

    If you find that to be a bit more of a mathematical slog than you're up for, google for "Schrodinger's equation infinite square well". That's a somewhat simpler example of how integer-only quantum numbers appear in a solution to Schrodinger's equation - it's the one that most introductory quantum mechanics textbooks start with. However, you'll still need at least a nodding acquaintance with differential equations.... without that we can't give you a better answer than mfb's answer above.
     
  5. Jul 16, 2015 #4
    In quantum mechanics a quantum number is "label" to address an eigenvalue of some conserved quantity (that in QM formalism is an operator that commutes with the hamiltonian), and the principale ones, for the classic problem of an electon that orbits around a nucleous, are the eigenvalues of the Hamiltonian itself (the energy), of the angular momentum, the magnetization and the spin operator.

    So it makes sanse to say that n (the principal quantum number) is equal to 1, since it says that the electon is in the first excited state, so the second smallest eigenvalue of the hamiltonian (which will have his own value in some unit of misure of the energy). On the contrary, it makes no sense at all to say that n=3/2 or some other fraction since it's simply a label from the lower state (the Hamiltonian must always be bound from below) to the infinity.
     
  6. Jul 16, 2015 #5
    The electron state in the atom is described by a probability wave, distributed around the nucleus. The wave is changing over time in a way specified by the Schrodinger equation. The point is not every function around the nucleus changing over time is consistent with the Schrodinger equation. Instead only a countable many functions are consistent with it. countable many means you can name all the solutions as F1, F2,F3.. The label becomes the principle quantum number
     
  7. Jul 17, 2015 #6

    Nugatory

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    That's true as far as it goes, but it doesn't explain why the energy eigenvalues must be discrete instead of forming a continuous spectrum as they do for an unbound electron. For that you have to solve Schrodinger's equation for the specific potential that we're desling with (##1/r## in this case).
     
  8. Jul 17, 2015 #7

    stevendaryl

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    Here's a way to think about quantum numbers that halfway explains them, but leaves a different mystery about them.

    For a typical one-dimensional problem (and if there is spherical symmetry, we can factor the wavefunction into a radial wave function and an angular wave function--the radial wave function is essentially a one-dimensional problem), we can write Schrodinger's equation this way:

    [itex]\frac{\partial^2 \psi}{\partial r^2} = \frac{-2m}{\hbar^2} (E - V) \psi[/itex]

    In a typical problem we have two regions:
    1. Oscillatory region, where [itex]E - V \gt 0[/itex]
    2. Non-oscillatory region, where [itex]E - V \lt 0[/itex]
    When [itex]E - V \gt 0[/itex], we can define [itex]k(r)[/itex] to be [itex]\sqrt{\frac{2m}{\hbar^2} (E-V)}[/itex]. In that region, the equation looks like:

    [itex]\frac{\partial^2 \psi}{\partial r^2} = -k^2(r) \psi[/itex]

    That equation has oscillatory solutions, like sines and cosines (it would be exactly sines and cosines if [itex]k[/itex] were constant). An oscillatory function goes up and down, and can be characterized by a nonnegative integer, namely, the number of zeros--the number of points where [itex]\psi(r) = 0[/itex]. That integer is the radial quantum number, [itex]n_r[/itex].

    That's an explanation for why the radial quantum number must be an integer. It doesn't really explain why energy levels are quantized, though. If you tried to numerically solve Schrodinger's equation for an arbitrary energy [itex]E[/itex], you would definitely get an integer number of zeros, no matter what the energy was. But what's special about energy eigenvalues can only be understood by looking at the other region--the non-oscillatory region.

    In this region, we have [itex]E - V \lt 0[/itex]. We can define [itex]\kappa(r)[/itex] to be [itex]\sqrt{\frac{2m}{\hbar^2} (V-E)}[/itex] In that region, the equation looks like:

    [itex]\frac{\partial^2 \psi}{\partial r^2} = +\kappa^2(r) \psi[/itex]

    Rather than oscillatory solutions, this equation has solutions that are like exponentials, they either blow up as [itex]r \rightarrow \infty[/itex], or they go to zero. If [itex]\kappa[/itex] were constant, then these two solutions would be [itex]e^{+\kappa r}[/itex] and [itex]e^{-\kappa r}[/itex]. This is where the quantization of energy levels comes from. If you try to solve the Schrodinger equation for an arbitrary energy level, what you'll find is that if you make sure that it is finite at [itex]r=0[/itex], then it will blow up as [itex]r \rightarrow \infty[/itex]. If you try to make it go to zero as [itex]r \rightarrow \infty[/itex], then the wave function will blow up at [itex]r=0[/itex]. Only for very special energies will it be possible to have the wave function be well behaved both at [itex]r=0[/itex] and at [itex]r \rightarrow \infty[/itex].

    What's not obvious from my explanation, and I think that the mathematics of proving this is beyond me, is the connection between those two different integral numbers. Why is there exactly one energy [itex]E_0[/itex] that has one zero and has nice behavior as [itex]r \rightarrow \infty[/itex] and [itex]r \rightarrow 0[/itex]?
     
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