34
$\begingroup$

I have been taught that the boundary conditions are just as important as the differential equation itself when solving real, physical problems.

When the Schrödinger equation is applied to the idealized hydrogen atom it is separable and boundary conditions are applied to the radial component. I am worried about the $r=0$ boundary near the nucleus. Near the proton, the electron's kinetic energy will be relativistic and looking at the Schrödinger equation itself for how this boundary should behave seems dangerous because its kinetic energy term is only a non-relativistic approximation.

Is there any physical intuition, or any math, that I can look at that should make me comfortable with the boundary condition in this region?

Improve this question
$\endgroup$
0

4 Answers 4

Reset to default
21
$\begingroup$

In solving the Schroedinger radial equation there is no boundary condition applied at $r=0$. At $r=\infty$ yes, $R(r)$ must tend to zero - so we reject the positive exponential solution; any change in that would have massive consequences. But there is no constraint laid on $R(r)$ or indeed $R'(r)$ as $r \to 0$.

So there's not a change in the boundary condition. There is a change in the kinetic and potential energies due to relativistic effects and the fact that the proton is not a point charge. These do have an effect - but very small, as the volume concerned is about $10^{-15}$ of the volume of the atom. (Actually atomic physicists experiments can detect these effects, at least for large $Z$ atoms, thanks to some very clever and precise optical experiments.) But this is a small effect, not the game changer that a new boundary condition could give.

Improve this answer
$\endgroup$
9
  • $\begingroup$ So, imagine that I take a numerical approach to solving the radial differential equation with $l=0$. I will "shoot" from $r=\infty$ inwards at various energy levels seeking the several eigenvalues for the s orbitals. If there is no boundary condition at $r=0$ then how do I know which energy levels are solutions? $\endgroup$
    – Paul Young
    Jun 2, 2019 at 14:43
  • 8
    $\begingroup$ Hint: trivially detectable for gold. Non-relativistic quantum mechanics yields a grey gold not a yellow gold. $\endgroup$
    – Joshua
    Jun 2, 2019 at 18:09
  • 3
    $\begingroup$ @PaulYoung try taking the limit of $\varepsilon\to0$ with potential being $\frac1{\sqrt{r^2+\varepsilon^2}}$ instead of the usual $1/r$. This will give you the result without imposing a special boundary condition. See also this question of mine and my self-answer there. $\endgroup$
    – Ruslan
    Jun 2, 2019 at 20:22
  • 1
    $\begingroup$ @PaulYoung I'm not sure what you're really asking: in the second paragraph of your OP you're speaking about kinetic energy operator being a non-relativistic approximation, which is irrelevant to boundary conditions (to approximate a bit closer you would use a higher order approximation of $$T=\sqrt{m^2c^4+p^2c^2}-mc^2=\frac{p^2}{2m}-\frac{p^4}{8m^3c^2}+\frac{p^6}{16m^5c^4}+...,$$ not change a boundary condition) and in the last paragraph you're asking about the boundary condition at $r=0$, which seems to be addressed by my link in the previous comment. $\endgroup$
    – Ruslan
    Jun 3, 2019 at 5:03
  • 1
    $\begingroup$ What I think @PaulYoung is getting at (and which I agree with) is that the statement "there is no constraint on $R(r)$ or $R'(r)$ as $r \to 0$." The wavefunction has to be finite as $r \to 0$. This effectively acts as a boundary condition for the radial ODE, since most solutions to the radial ODE are not finite at the origin. $\endgroup$
    – Michael Seifert
    Jun 3, 2019 at 11:40
16
$\begingroup$

Good question. Your claim that

near the proton the electron's kinetic energy will be relativistic

is not as straightforward as it might seem. The electron's kinetic energy $\langle \hat{T} \rangle = \langle \hat{p}^2 \rangle / (2m)$ is a nonlocal quantity that can be equivalently expressed as either of the two integrals

$$\langle \hat{T} \rangle = \frac{1}{2m} \int d^3x\ \psi^*(x) \left(-\hbar^2 \nabla^2 \right) \psi(x) = \frac{1}{2m} \int d^3x\ |\hbar\, {\bf \nabla} \psi(x)|^ 2.$$

So the electron's kinetic energy "at" a particular location is not well-defined; it could be the value of either of the two integrands above at that point (or, indeed, of any other integrand that integrates to the same value over all of space).

The latter expression is the more natural one to use, though, because at least it's positive-semidefinite. We still have the problem that $\hbar^2 |\nabla \psi(0)|^2/(2m)$ is a "kinetic energy density" (whatever that is) rather than an actual kinetic energy, so we can't speak of how relativistic the electron is "at" the nucleus. (We could integrate over the empirical size of the nucleus, but I don't think that's really what your question is getting at - you're not asking about when the electron is literally inside the nucleus, but when it's close enough to the potential center that it's intuitively moving very quickly.)

But none of this really matters - the point is that since the integrand is positive-definite, the contribution to the kinetic energy over any particular region is always less than (or equal to) the total kinetic energy over every region. So to meaningfully check whether relativistic effects need to be taken into account, you need to calculate the total kinetic energy over all of space. This turns out to be $\hbar^2/(2m a^2) = m e^4/(2 \hbar^2) = (\alpha^2 /2) m c^2$, where $\alpha$ is the fine-structure constant. Relativistic effects are negligible if the kinetic energy is much less than the electron's rest energy, which corresponds to the condition that $\alpha^2/2 = 1/37538 \ll 1$, which, reassuringly, is true.

Improve this answer
$\endgroup$
9
  • 1
    $\begingroup$ I understand from this that the electron's average or "expected" kinetic energy is not relativistic. However, the Schrödinger equation looks "local" to me and I am kinda thinking it does seem reasonable to think about it having a value at every point. This is what makes the wave function "wigglier" near the nucleus, right? $\endgroup$
    – Paul Young
    Jun 2, 2019 at 14:37
  • 2
    $\begingroup$ Whether it's wigglier near the nucleus depends on the orbital - for some choices of $n$ and $l$, the wave function is flat at the nucleus. Anyway, how do you propose to covert the kinetic energy density given by the wave function's gradient to an actually kinetic energy, which will tell you whether the electron is relativistic? $\endgroup$
    – tparker
    Jun 2, 2019 at 17:45
  • $\begingroup$ I propose to locally apply the momentum operator and divide the result by the rest mass and compare it to the speed of light $\endgroup$
    – Paul Young
    Jun 2, 2019 at 18:12
  • 2
    $\begingroup$ If you do that then the units don't work out. That's what I keep trying to tell you; the momentum operator only returns a quantity with the units of momentum if you integrate it over space. (Also, del is "\nabla" in TeX.) $\endgroup$
    – tparker
    Jun 2, 2019 at 18:50
  • 2
    $\begingroup$ @PaulYoung The physical significance of that ratio isn't clear to me. It would be very large at any location where the electron is very unlikely to be, and I don't see why that would correspond to the kinetic energy "at" that location. I just don't think there's any natural local way to define the kinetic energy "at" a location within the setup of nonrelativistic QM. (In the QFT setting, you can talk about the energy-momentum tensor.) $\endgroup$
    – tparker
    Jun 3, 2019 at 17:36
8
$\begingroup$

The boundary conditions that pick up the hydrogen wave functions are the "constraints" placed on the wave function solutions. Remember that the observable is the probability distribution from the $Ψ^*Ψ$, not a particular location. Please read the link. The solutions are within the quantum mechanic postulates after all.

There isn't any relativistic boundary condition, because there are no orbits, only probability distributions.

Thus the solutions do not have a singularity at r=0, and in general there is a small probability of finding the electron at the origin, if the quantum numbers allow for an interaction, as with electron capture in nuclei. For the hydrogen atom there is not enough energy for a neutron to appear.

Improve this answer
$\endgroup$
5
  • $\begingroup$ I am a tad bit worried about the GSU links because they seem to be based on an assumption of non-relativity, but that is exactly what I am worried about $\endgroup$
    – Paul Young
    Jun 1, 2019 at 16:15
  • $\begingroup$ But the same solutions as far as energy levels etc come out of solving the hydrogen atom with the Dirac equation, which is relativistic. quantummechanics.ucsd.edu/ph130a/130_notes/node501.html . This is due to the probabilistic nature of quantum mechanics, where the constraints and the postulates are the ones determning the solutions. There is no effect of "nearing r=0" in the probability distributions in either case. $\endgroup$
    – anna v
    Jun 1, 2019 at 17:03
  • $\begingroup$ I don't understand the Dirac equation, but it is sounding from your answer and from Prof Barlow that since the $r=0$ boundary condition does not derive from anything that has to do with the equation, there isn't that much to worry about other than the small energy correction which should be on a "volume" scale $\endgroup$
    – Paul Young
    Jun 1, 2019 at 17:09
  • $\begingroup$ the link I give above says "This result gives the same answer as our non-relativistic calculation to order a^4 but is also correct to higher order. It is an exact solution to the quantum mechanics problem posed but does not include the effects of field theory, such as the Lamb shift and the anomalous magnetic moment of the electron. " $\endgroup$
    – anna v
    Jun 1, 2019 at 17:10
  • $\begingroup$ challenging looking link, but I see what I can do with it $\endgroup$
    – Paul Young
    Jun 1, 2019 at 17:37
6
$\begingroup$

The boundary condition at r=0 is that the wave function should be finite. The Schrödinger equation for hydrogen UC atoms and likely all atoms has solutions with negative $\cal l$, which are rejected because they diverge at r=0. See for example Schiff's textbook on quantum mechanics.

As for relativistic effects, you may want to compare the hydrogen energy expressions for Dirac, better, Klein-Gordon - no spin, and Schrödinger. Check out another great text, Itzykson and Zuber, for these .

Improve this answer
$\endgroup$
1
  • $\begingroup$ I will check out the reference texts. Schiff's, in particular, sounds like there are some comments I have never seen. $\endgroup$
    – Paul Young
    Jun 2, 2019 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.