Degenerate States in Quantum Mechanics

In his book on quantum mechanics in the chapter on perturbation theory Dirac says in a footnote:

A system with only one stationary state belonging to each energy-level is often called non-degenerate and one with two or more stationary states belonging to an energy-level is called degenerate, although these words are not very appropriate from the modern point of view.

I have two questions about this:

1) Why did Dirac deem the terms (non-)degenerate inappropriate?

2) Why do we, with our even more modern point of view, still use them?

Stability of a Degenerate Equilibrium Point in a Planar ODE

Consider the planar ODE

$dot x_1 = x_2$

$dot x_2 = – x_1^2 – 2 x_1 – 1$

Obliviously, $(x_1,x_2)=(-1,0)$ is an equilibrium point. The Jacobian matrix at this point is

$$J = begin{bmatrix}
0 & 1 \
0 & 0
end{bmatrix}$$

Thus, linearizarion fails in determining the stability. How can we determine the stability of this equilibrium point?

Degenerate States in Quantum Mechanics

In his book on quantum mechanics in the chapter on perturbation theory Dirac says in a footnote:

A system with only one stationary state belonging to each energy-level is often called non-degenerate and one with two or more stationary states belonging to an energy-level is called degenerate, although these words are not very appropriate from the modern point of view.

I have two questions about this:

1) Why did Dirac deem the terms (non-)degenerate inappropriate?

2) Why do we, with our even more modern point of view, still use them?

computation of an integral for 2nd order non degenerate perturbation theory

I am given that the potential of a diatomic molecule is equal to $$V(rho)=-2V left ( frac{1}{rho ^2}-frac{1}{2 rho ^2} right ) $$ With $rho=r/a$ is a dimesionless coordinate, and $r$ is the separation distance between the two atoms. I found the first order corrections without issue, but I am stuck on finding the second order one. I know that
$$E_n^2=sum_{mneq n}frac{|langlepsi_m^0|H’|psi_n^0rangle|^2}{E_n^0-E_m^0}$$
For the given problem, I also found that the wavefunctions are $psi=e^{-x^2/2} H_{n}(x)$ (hermite polynomials). My problem is in computing the integral of the inner product, i.e, computing
$$langlepsi_m^0|H’|psi_n^0rangle=int_{-infty}^{infty} x^3 e^{-x^2} H_{n}(x)H_{m}(x)dx$$
I could apply integration by parts a bazillion times to obtain the answer, but it is far too tedious. From reading Griffiths, I know that there is a way to do this much simpler with dirac notation and ladder operators, in conjunction with the usual ladder operator identities, but I am unsure on how to go about this.

What is $mathrm{O}_q/mathrm{SO}_q$ if $q$ is a quadratic $mathbb{Z}$-form which is degenerate?

Any binary quadratic $mathbb{Z}$-form $q$ induces a symmetric bilinear form
$$ B_q(u,v) = q(u+v) – q(u) -q(v) forall u,v inmathbb{Z}^2 $$
and it is considered non-degenerate (over $mathbb{Z}$) if its discriminant
$$ text{disc}(q) := det(B_q(e_i,e_j)_{1 leq i,j leq 2}) $$
where $e_1 = (1,0)$ and $e_2 = (0,1)$ is invertible in $mathbb{Z}$, i.e., equals $pm 1$: see (2.1) in: http://math.stanford.edu/~conrad/papers/redgpZsmf.pdf .
Suppose $q$ is degenerate, but still $text{disc}(q) neq 0$ (so it is non-degenerate over $mathbb{Q}$). So its special orthogonal group scheme $SO_q$ defined over $text{Spec} mathbb{Z}$, does not have to be smooth, but it is flat as $mathbb{Z}$ is Dedekind (loc. sit. Definition 2.8 and right after), and it is closed in the full orthogonal group $O_q$, whence the quotient $Q:=O_q/SO_q$ is representable.

My question is: Is $Q$ a finite group of order $2$ over $text{Spec} mathbb{Z}$ ?
Apparently, when applied to any integral domain $R$ which is an extension of $mathbb{Z}$, the elements of $O_q(R)$, in some matrix realization, must be of $det pm 1$, so we could think of $Q$ as a $mathbb{Z}$-group of order $2$, but as a functor of points, $O_q$ can be applied to any $mathbb{Z}$-algebra $R$, for which we may find elements of $O_q(R)$ which are not of $det = pm 1$.

For example, let $q(x,y)=x^2+y^2$. One can verify it is degenerate.
We get $SO_q = text{Spec} mathbb{Z}[x,y]/(x^2+y^2-1)$. Consider its matrix realization $$ left { A=left( begin{array}{cc}
x & -y \
y & x \
end{array}right): det(A)=1 right }. $$

Then the component of $det = -1$ elements in $O_q$ is obtained by $text{diag}(1,-1)SO_q$.
So apparently, $Q = mu_2$ (which unlike the other order $2$ group $(mathbb{Z}/2)_mathbb{Z}$, it has a double point at the reduction at $(2)$, not two distinct ones).
So far everything is good. But $A=text{diag}(3,1)$ belongs to $O_q(R)$ where $R=mathbb{Z}/8$ (as $A^T cdot A = I_2$ in $R$, where $I_2$ represents $q$), but $det(A)neq pm 1$ in $R$ !
Does it mean that $O_q$ is more than these two connected components ?

I thought to avoid this problem by considering $O_q$ and $SO_q$ as flat sheaves (in the small site of flat extensions of $mathbb{Z}$, since what I really care is of $H^1_text{fppf}(mathbb{Z},O_q)$), but we may still find extensions such as $mathbb{Z} times mathbb{Z}$ containing a square root of unity other than $-1$ ?!

Could you please help ?

Thank you !

Rony

Degenerate States in Quantum Mechanics

In his book on quantum mechanics in the chapter on perturbation theory Dirac says in a footnote:

A system with only one stationary state belonging to each energy-level is often called non-degenerate and one with two or more stationary states belonging to an energy-level is called degenerate, although these words are not very appropriate from the modern point of view.

I have two questions about this:

1) Why did Dirac deem the terms (non-)degenerate inappropriate?

2) Why do we, with our even more modern point of view, still use them?

Degenerate States in Quantum Mechanics

In his book on quantum mechanics in the chapter on perturbation theory Dirac says in a footnote:

A system with only one stationary state belonging to each energy-level is often called non-degenerate and one with two or more stationary states belonging to an energy-level is called degenerate, although these words are not very appropriate from the modern point of view.

I have two questions about this:

1) Why did Dirac deem the terms (non-)degenerate inappropriate?

2) Why do we, with our even more modern point of view, still use them?

Degenerate States in Quantum Mechanics

In his book on quantum mechanics in the chapter on perturbation theory Dirac says in a footnote:

A system with only one stationary state belonging to each energy-level is often called non-degenerate and one with two or more stationary states belonging to an energy-level is called degenerate, although these words are not very appropriate from the modern point of view.

I have two questions about this:

1) Why did Dirac deem the terms (non-)degenerate inappropriate?

2) Why do we, with our even more modern point of view, still use them?

Neuton star (or any degenerate matter) and length contraction

I know there are plenty of questions abound on the internet that go something like this:

“If an object were to move fast enough, would it collapse into a black hole”?

And the (correct) answer that is often given in response: “No, that would be an example of relativistic mass, and it would not match predictions/calculations.”

But I have to ask, just to “drive it home” intuitively: let’s say we had degenerate matter, let’s call it a neutron star. This star was moving horizontally relative to me (so not spinning). Even in that case, would I observe it to length contract or collapse into a black hole?

It would length contract, right?