## 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.

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.

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?

## Showing that a 2-form on an odd dimensional space is not degenerate

On an odd-dimensional space $$mathbb R^{2n+1}$$ with coordinates $$x_1…x_n;y_1…y_n;t$$ consider the following 2-form:
$$domega=sum dx_i land dy_i-omega land dt$$
where $$omega$$ is any 1-form on $$mathbb R^{2n+1}$$.
How to show that $$domega$$ is non degenerate?

## 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 $$-1$$ ?!

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.

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.

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.

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?