- #1
HJ Farnsworth
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Greetings everyone,
I'm going through Griffiths QM, and just wanted to make sure I understood the roles (where they come from, etc.) of the quantum numbers n, l, and m. What is written below is basically an outline of my current understanding, of which I am seeking either a confirmation or correction of (please forgive any clumsy equations, I don't have my book with me and don't feel like jumping around the internet too much). At the end of this post is a quick summary of my interpretation of the quantum numbers and the actual questions I have.
Principle quantum number, n:
Start with the TDSE. Attempt to solve using separation of variables:
[itex]\Psi[/itex](r,t)=[itex]\psi[/itex](r)f(t)
If the potential is such that this separation meets the boundary condition of normalizability, then there will be infinitely many solutions (due basically to the fact that increasing the periodicity of the function can result in different solutions that are normalizable, eg., particle in a box), labelled by the principle quantum number, n. Each solution corresponds to a state of definite total energy, and [itex]\Psi[/itex] is the sum of each [itex]\psi[/itex] multiplied by the probability of that [itex]\psi[/itex]:
[itex]\Psi[/itex](r,t)=[itex]\sum[/itex]cn[itex]\psi[/itex]n(r)f(t)
If, on the other hand, the potential is such that separation does not meet the boundary condition, then there are no states of definite total energy, the solutions are labelled by the continuous variable k=[itex]\sqrt{2mE}[/itex]/[itex]\hbar[/itex]. Instead of cn, a probability density function is used, and [itex]\Psi[/itex] is:
(1/[itex]\sqrt{2π}[/itex])∫[itex]\phi[/itex](k)[itex]\psi[/itex]n(r)f(t)dk
Azimuthal and magnetic quantum numbers, l and m:
Start with the TISE. Assume spherically symmetric potential and attempt to solve using separation of variables:
[itex]\psi[/itex](r,[itex]\theta[/itex],[itex]\varphi[/itex])=R(r)Y([itex]\theta[/itex],[itex]\varphi[/itex])
Eventually, this turns into an equation of the form f(r)=g([itex]\theta[/itex],[itex]\varphi[/itex]), so that both are equal to a constant, which we label l(l+1). Only the radial part of the equation has an energy term or a potential term, so its solution (and whether or not it turns out that doing separation of variables results in a legal solution) depends on the potential V(r). Assuming it does meet the boundary conditions, the radial solution is now dependent on two quantum numbers, so energy is still E=En, but now:
R(r)=Rnl(r)
For the angular equation, attempt separation of variables once more:
Y([itex]\theta[/itex],[itex]\varphi[/itex])=[itex]\Theta[/itex]([itex]\theta[/itex])[itex]\Phi[/itex]([itex]\varphi[/itex])
As before, this turns into an equation of the form h([itex]\theta[/itex])=j([itex]\varphi[/itex]), so that both are equal to a constant, which we label m2. Solving the two differential equations, the solution to the total angular equation is the spherical harmonics, which from the way we labelled the separation constants, depend on both quantum numbers l and m:
Y([itex]\theta[/itex],[itex]\varphi[/itex])=Y[itex]^{m}_{l}[/itex]([itex]\theta[/itex],[itex]\varphi[/itex])
So the solution to the TISE depends on all three quantum numbers:
[itex]\psi[/itex][itex]^{m}_{nl}[/itex](r,[itex]\theta[/itex],[itex]\varphi[/itex])=Rnl(r)Y[itex]^{m}_{l}[/itex]([itex]\theta[/itex],[itex]\varphi[/itex])
Bringing angular momentum into the picture, it turns out that solutions to the TISE that are separable as described above represent states of definite L2 and Lz (so basically states where the total angular momentum and one of its components can be known with certainty):
[itex]\widehat{H}[/itex][itex]\psi[/itex][itex]^{m}_{nl}[/itex]=E[itex]\psi[/itex][itex]^{m}_{nl}[/itex]
[itex]\widehat{L}[/itex]2[itex]\psi[/itex][itex]^{m}_{nl}[/itex]=[itex]\hbar[/itex]2l(l+1)[itex]\psi[/itex][itex]^{m}_{nl}[/itex]
[itex]\widehat{L}[/itex]z[itex]\psi[/itex][itex]^{m}_{nl}[/itex]=[itex]\hbar[/itex]m[itex]\psi[/itex][itex]^{m}_{nl}[/itex]
Mirroring the principle quantum number construction, the wavefunction for a given energy level can be written as a summation over l and m of each [itex]\psi[/itex][itex]^{m}_{nl}[/itex], multiplied by the probability of being at the L2 labeled by l and the Lz labeled by m:
[itex]\psi[/itex]n=[itex]\sum[/itex]lmc[itex]^{m}_{l}[/itex]Rnl(r)Y[itex]^{m}_{l}[/itex]([itex]\theta[/itex],[itex]\varphi[/itex])
SUMMARY
[itex]\Psi[/itex](r,t)=[itex]\psi[/itex](r)f(t)[itex]\Rightarrow[/itex][itex]\widehat{H}[/itex][itex]\psi[/itex]n=E[itex]\psi[/itex]n, a state of definite total energy, indexed by quantum number n.
[itex]\psi[/itex](r,θ,[itex]\varphi[/itex])=R(r)Y([itex]\theta[/itex],[itex]\varphi[/itex])[itex]\Rightarrow[/itex][itex]\widehat{L}[/itex]2[itex]\psi[/itex]nl=[itex]\hbar[/itex]2l(l+1)[itex]\psi[/itex]nl, a state of definite L2, so basically a state of definite total angular momentum, indexed by quantum number l.
Y([itex]\theta[/itex],[itex]\varphi[/itex])=[itex]\Theta[/itex]([itex]\theta[/itex])[itex]\Phi[/itex]([itex]\varphi[/itex])[itex]\Rightarrow[/itex][itex]\widehat{L}[/itex]z[itex]\psi[/itex][itex]^{m}_{nl}[/itex]=[itex]\hbar[/itex]m[itex]\psi[/itex][itex]^{m}_{nl}[/itex], a state of definite Lz, so basically a state where one component of the angular momentum is known with certainty, indexed by quantum number m.
QUESTIONS
1. What is k=[itex]\sqrt{2mE}[/itex]/[itex]\hbar[/itex] called (it seems like there must be a name for it)?
2. If the potential is such that [itex]\psi[/itex](r,[itex]\theta[/itex],[itex]\varphi[/itex])=R(r)Y([itex]\theta[/itex],[itex]\varphi[/itex]) ends up not satisfying the boundary condition (ie., a state of definite total energy without definite total angular momentum), what is the integral mirroring (1/[itex]\sqrt{2π}[/itex])∫[itex]\phi[/itex](k)[itex]\psi[/itex]n(r)f(t)dk, and what is the variable mirroring k? So basically, could you please rewrite the equation immediately preceding the summary in what would be the appropriate integral form?
3. If the equation [itex]\psi[/itex](r,[itex]\theta[/itex],[itex]\varphi[/itex])=R(r)Y([itex]\theta[/itex],[itex]\varphi[/itex]) satisfies the boundary condition, are there situations where separating the angular equation Y([itex]\theta[/itex],[itex]\varphi[/itex])=[itex]\Theta[/itex]([itex]\theta[/itex])[itex]\Phi[/itex]([itex]\varphi[/itex]) will still end up being illegal (ie., are there situations where the total angular momentum can be known with certainty, but none of its components can?) If so, repeat question 2 with the angular separation in place of the TISE separation.
4. This is similar to question 3. In the 2nd and 3rd equations in my summary, I'm saying that as I understand it, if separating the TISE results in legal equations for the wavefunction, then the total angular momentum is known with certainty regardless to whether separating the angular equation results in legal equations for the wavefunction, whereas to know a component of L, you have to be able to separate the angular equation. Is this correct, or do you have to be able to separate both legally to know L2 with certainty?
Thanks for any help you can give.
-HJ Farnsworth
I'm going through Griffiths QM, and just wanted to make sure I understood the roles (where they come from, etc.) of the quantum numbers n, l, and m. What is written below is basically an outline of my current understanding, of which I am seeking either a confirmation or correction of (please forgive any clumsy equations, I don't have my book with me and don't feel like jumping around the internet too much). At the end of this post is a quick summary of my interpretation of the quantum numbers and the actual questions I have.
Principle quantum number, n:
Start with the TDSE. Attempt to solve using separation of variables:
[itex]\Psi[/itex](r,t)=[itex]\psi[/itex](r)f(t)
If the potential is such that this separation meets the boundary condition of normalizability, then there will be infinitely many solutions (due basically to the fact that increasing the periodicity of the function can result in different solutions that are normalizable, eg., particle in a box), labelled by the principle quantum number, n. Each solution corresponds to a state of definite total energy, and [itex]\Psi[/itex] is the sum of each [itex]\psi[/itex] multiplied by the probability of that [itex]\psi[/itex]:
[itex]\Psi[/itex](r,t)=[itex]\sum[/itex]cn[itex]\psi[/itex]n(r)f(t)
If, on the other hand, the potential is such that separation does not meet the boundary condition, then there are no states of definite total energy, the solutions are labelled by the continuous variable k=[itex]\sqrt{2mE}[/itex]/[itex]\hbar[/itex]. Instead of cn, a probability density function is used, and [itex]\Psi[/itex] is:
(1/[itex]\sqrt{2π}[/itex])∫[itex]\phi[/itex](k)[itex]\psi[/itex]n(r)f(t)dk
Azimuthal and magnetic quantum numbers, l and m:
Start with the TISE. Assume spherically symmetric potential and attempt to solve using separation of variables:
[itex]\psi[/itex](r,[itex]\theta[/itex],[itex]\varphi[/itex])=R(r)Y([itex]\theta[/itex],[itex]\varphi[/itex])
Eventually, this turns into an equation of the form f(r)=g([itex]\theta[/itex],[itex]\varphi[/itex]), so that both are equal to a constant, which we label l(l+1). Only the radial part of the equation has an energy term or a potential term, so its solution (and whether or not it turns out that doing separation of variables results in a legal solution) depends on the potential V(r). Assuming it does meet the boundary conditions, the radial solution is now dependent on two quantum numbers, so energy is still E=En, but now:
R(r)=Rnl(r)
For the angular equation, attempt separation of variables once more:
Y([itex]\theta[/itex],[itex]\varphi[/itex])=[itex]\Theta[/itex]([itex]\theta[/itex])[itex]\Phi[/itex]([itex]\varphi[/itex])
As before, this turns into an equation of the form h([itex]\theta[/itex])=j([itex]\varphi[/itex]), so that both are equal to a constant, which we label m2. Solving the two differential equations, the solution to the total angular equation is the spherical harmonics, which from the way we labelled the separation constants, depend on both quantum numbers l and m:
Y([itex]\theta[/itex],[itex]\varphi[/itex])=Y[itex]^{m}_{l}[/itex]([itex]\theta[/itex],[itex]\varphi[/itex])
So the solution to the TISE depends on all three quantum numbers:
[itex]\psi[/itex][itex]^{m}_{nl}[/itex](r,[itex]\theta[/itex],[itex]\varphi[/itex])=Rnl(r)Y[itex]^{m}_{l}[/itex]([itex]\theta[/itex],[itex]\varphi[/itex])
Bringing angular momentum into the picture, it turns out that solutions to the TISE that are separable as described above represent states of definite L2 and Lz (so basically states where the total angular momentum and one of its components can be known with certainty):
[itex]\widehat{H}[/itex][itex]\psi[/itex][itex]^{m}_{nl}[/itex]=E[itex]\psi[/itex][itex]^{m}_{nl}[/itex]
[itex]\widehat{L}[/itex]2[itex]\psi[/itex][itex]^{m}_{nl}[/itex]=[itex]\hbar[/itex]2l(l+1)[itex]\psi[/itex][itex]^{m}_{nl}[/itex]
[itex]\widehat{L}[/itex]z[itex]\psi[/itex][itex]^{m}_{nl}[/itex]=[itex]\hbar[/itex]m[itex]\psi[/itex][itex]^{m}_{nl}[/itex]
Mirroring the principle quantum number construction, the wavefunction for a given energy level can be written as a summation over l and m of each [itex]\psi[/itex][itex]^{m}_{nl}[/itex], multiplied by the probability of being at the L2 labeled by l and the Lz labeled by m:
[itex]\psi[/itex]n=[itex]\sum[/itex]lmc[itex]^{m}_{l}[/itex]Rnl(r)Y[itex]^{m}_{l}[/itex]([itex]\theta[/itex],[itex]\varphi[/itex])
SUMMARY
[itex]\Psi[/itex](r,t)=[itex]\psi[/itex](r)f(t)[itex]\Rightarrow[/itex][itex]\widehat{H}[/itex][itex]\psi[/itex]n=E[itex]\psi[/itex]n, a state of definite total energy, indexed by quantum number n.
[itex]\psi[/itex](r,θ,[itex]\varphi[/itex])=R(r)Y([itex]\theta[/itex],[itex]\varphi[/itex])[itex]\Rightarrow[/itex][itex]\widehat{L}[/itex]2[itex]\psi[/itex]nl=[itex]\hbar[/itex]2l(l+1)[itex]\psi[/itex]nl, a state of definite L2, so basically a state of definite total angular momentum, indexed by quantum number l.
Y([itex]\theta[/itex],[itex]\varphi[/itex])=[itex]\Theta[/itex]([itex]\theta[/itex])[itex]\Phi[/itex]([itex]\varphi[/itex])[itex]\Rightarrow[/itex][itex]\widehat{L}[/itex]z[itex]\psi[/itex][itex]^{m}_{nl}[/itex]=[itex]\hbar[/itex]m[itex]\psi[/itex][itex]^{m}_{nl}[/itex], a state of definite Lz, so basically a state where one component of the angular momentum is known with certainty, indexed by quantum number m.
QUESTIONS
1. What is k=[itex]\sqrt{2mE}[/itex]/[itex]\hbar[/itex] called (it seems like there must be a name for it)?
2. If the potential is such that [itex]\psi[/itex](r,[itex]\theta[/itex],[itex]\varphi[/itex])=R(r)Y([itex]\theta[/itex],[itex]\varphi[/itex]) ends up not satisfying the boundary condition (ie., a state of definite total energy without definite total angular momentum), what is the integral mirroring (1/[itex]\sqrt{2π}[/itex])∫[itex]\phi[/itex](k)[itex]\psi[/itex]n(r)f(t)dk, and what is the variable mirroring k? So basically, could you please rewrite the equation immediately preceding the summary in what would be the appropriate integral form?
3. If the equation [itex]\psi[/itex](r,[itex]\theta[/itex],[itex]\varphi[/itex])=R(r)Y([itex]\theta[/itex],[itex]\varphi[/itex]) satisfies the boundary condition, are there situations where separating the angular equation Y([itex]\theta[/itex],[itex]\varphi[/itex])=[itex]\Theta[/itex]([itex]\theta[/itex])[itex]\Phi[/itex]([itex]\varphi[/itex]) will still end up being illegal (ie., are there situations where the total angular momentum can be known with certainty, but none of its components can?) If so, repeat question 2 with the angular separation in place of the TISE separation.
4. This is similar to question 3. In the 2nd and 3rd equations in my summary, I'm saying that as I understand it, if separating the TISE results in legal equations for the wavefunction, then the total angular momentum is known with certainty regardless to whether separating the angular equation results in legal equations for the wavefunction, whereas to know a component of L, you have to be able to separate the angular equation. Is this correct, or do you have to be able to separate both legally to know L2 with certainty?
Thanks for any help you can give.
-HJ Farnsworth