I have been reading "Introduction to Quantum Mechanics by David J. Griffiths" and in the chapter 6- Time-Independent Perturbation theory, when we are explaining the fine structure via relativistic correction section 6.3.1) and spin-orbit coupling(6.3.2), we find that the states with $\ell=0$ are often problematic, or in other words, the general formula fails for them. For example:

(a) In the relativistic correction, we find that the general formula we derived doesn't quite work for states with $\ell=0$ because for these states $p^2$ is hermitian while $p^4$ is not, where $p$=momentum operator.

(b) And in spin orbit coupling, we find that general formula doesn't work for $\ell=0$ states as it leads to an indeterminate form.

I am aware about why this happens mathematically but I don't know what's physically there, that is if there is something special about $\ell=0$ states?

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    $\begingroup$ Hi Harshdeep, takin get screenshots of texts is generally looked down upon for a few reasons (e.g., doesn't allow for screen readers). It would be preferable for you to transcribe the text into the site using MathJax (search 'notation' in help center) $\endgroup$
    – Kyle Kanos
    Aug 11, 2019 at 11:27
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    $\begingroup$ (see here for more details, and here for a MathJax tutorial.) $\endgroup$ Aug 11, 2019 at 12:41
  • $\begingroup$ @KyleKanos okay i will take care of this in my future questions. Actually my question is complete all by itself, the screenshots are just supplement for anyone who want to see the exact formula. But yeah i get it i should've used mathjax $\endgroup$ Aug 11, 2019 at 13:14
  • $\begingroup$ Don't take care of this in future questions... take care of this NOW. $\endgroup$ Aug 11, 2019 at 13:52
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    $\begingroup$ Related: physics.stackexchange.com/q/460326 $\endgroup$ Aug 11, 2019 at 14:07

3 Answers 3


In the link provided by DanielC in the comments, a user has pointed out that the $l=0$ states have a kink at the origin i.e. the derivative(~$p$) is discontinuous. Now, roughly speaking the derivative of a discontinuous function is a delta function(~$p^2$), and the (double)derivative of the delta function(~$p^4$) is not a well-defined physical state. This is the reason why $p^4$ is not a hermitian operator, and thus won't have an observable value for $l=0$.

Now, you might wonder why $l=0$ states have a kink at the origin. If you look at the forms of the waves $R_{nl}$, you will see that only those with $l=0$ have a non-zero value at the origin. Roughly, then..these are the only states where the electron can get arbitrarily close to the nucleus. Now, the potential is infinite at the origin; and therefore so is $p^2$, which is a double derivative of the wavefunction. One way to have an infinite double derivative is to have a discontinuous single derivative. Thus the kink.

P.S. The infinitude of $p^2$ at the origin can also be roughly seen to be a reflection of the classical fact that you will need infinite energy to go to the nucleus. Thus, perturbation theory breaks down here. This also explains why Dirac's equation gives the exact result.

  • $\begingroup$ What I find amazing is that Griffiths's book is the only place I could find that $p^4$ is non self-adjoint in the subspace of solutions of SE for discrete spectrum, as it renders the whole machinery of all books to compute the perturbation due to special relativity as mathematically flawed (in strict terms, the conditions of Kato-Rellich theorem do not apply). $\endgroup$
    – DanielC
    Aug 13, 2019 at 18:59
  • $\begingroup$ I have found exactly why derivatives of delta functions don't form well-defined physical states, while the dirac delta function itself does. According to mathworld.wolfram.com/DeltaFunction.html, the $n^{th}$ derivative of the delta function is defined as: $\delta^{(n)}(x)=\frac{(-1)^{n}n!\delta (x)}{x^n}$. So for any derivative, the $\frac{1}{x^n}$ factor will yield infinite expectation values for many observables $\endgroup$
    – Mani Jha
    Aug 14, 2019 at 7:04
  • $\begingroup$ By the way, my argument indicates that $p^3$ should be non-hermitian too for these states. Could anyone check? $\endgroup$
    – Mani Jha
    Aug 14, 2019 at 7:21

I don't think there is anything physically interesting about $l=0$, it is just outside the basic premise of the mathematical model that was developed. I mean, if you develop a mathematical construction for spin-orbit coupling, you should not expect the model to work when your input is: there is no orbit. It happens a lot with mathematical constructs in physics and sometimes the math is nice enough to accommodate even the edge cases but that is not always the case.

Consider the Newton's second law. When a constant force $F$ acts on a (massles) box of $n$ particles of mass $m$ each, the box will accelerate at $$a=F/(nm).$$ For $n=0$ (or $m=0$) this model does not apply because: How do you apply a force on $n=0$ particles? Do $n=0$ particles accelerate to infinite velocity in 0 time for any $F>0$? Can $a$ really be infinite?


Note that $p^2$ is NOT an Hermitian MATRIX since.

$$p^2~~=~~ -\dfrac{\hbar^2}{r^2}\dfrac{d}{dr}\left( r^2\dfrac{d}{dr} \right) ~~=~~ -\hbar^2\left(\dfrac{2}{r}\dfrac{\partial}{\partial r}+\dfrac{\partial^2}{\partial r^2}\right)$$

It is the sum of (infinite dimensional) real matrices and the righthand side tells us that one of them is anti-symmetric $(\partial_r)$ while the other is symmetric $(\partial^2_r)$. The diagonals of these matrices look like:

$$ \begin{smallmatrix} 0&1&0&0&0&0&0 \\ -1&0&1&0&0&0&0 \\ 0&-1&0&1&0&0&0 \\ 0&0&-1&0&1&0&0 \\ 0&0&0&-1&0&1&0 \\ 0&0&0&0&-1&0&1 \\ 0&0&0&0&0&-1&0 \end{smallmatrix} ~~~~~~~~~~ \begin{smallmatrix} 2&-1&0&0&0&0&0&0 \\ -1&2&-1&0&0&0&0&0 \\ 0&-1&2&-1&0&0&0&0\\ 0&0&-1&2&-1&0&0&0 \\ 0&0&0&-1&2&-1&0&0 \\ 0&0&0&0&-1&2&-1&0 \\ 0&0&0&0&0&-1&2&-1 \\ 0&0&0&0&0&0&-1&2 \end{smallmatrix} $$

A real matrix must be symmetric to be Hermitian so it is the $\partial_r$ that destroys the symmetry. (The complete term $\tfrac{2}{r}\partial_r$ is not symmetric either).

Now in some cases it may be that it for fills the equation $\langle f \,|\, p^2\,g \rangle = \langle p^2\,f \,| \,g \rangle$ which is used as the definition of an Hermitian operator depending on $f$ and $g$. For instance the trivial $f=g=0$. There is no reason to assume that for less trivial examples of $f$ and $g$ this somehow should be true for $p^4$ as well.

Some good references:

Spherical Polar Laplacian: Quantum Physics, Eisberg & Resnick appendix M Hermitian Operators: Mathematical methods for Physicists, Chapter 10.2

Original text from the book:

enter image description here

  • $\begingroup$ But if this is in general true, how can we describe hermiticity of the Hamiltonian itself? Plus the proof that you have attached works for every well behaved function, only those functions which donot tend to fall to zero at infinity will cause any trouble. $\endgroup$ Aug 24, 2019 at 19:29

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