# Determining if an operator is degenerate

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• Physgeek64
In summary, the conversation discusses how to formally determine whether a given operator is degenerate. It is mentioned that degeneracy is often a result of symmetry, and in the example given, the operator is translation invariant, resulting in a set of mutually degenerate states. It is also mentioned that there may be other methods for finding degeneracies, but they are more advanced. The conversation concludes with a discussion on how to write the degeneracy equation for a differential operator.
Physgeek64
Hi,

I was wondering how you can formally determine whether a given operator is degenerate. I undertand you can produce the 'usual equation' det(Q-##\lambda ##)=0 and solve for ##\lambda ##, where Q is our operator. But if Q is a differential (for example ##\frac{p^2}{2m}= - \bar{h} \frac{d^2}{dx^2}## ) how can one write such an equation.

p.s. I only used this example as I know that the operator is degenerate with sin and cos of the same argument. Other examples are welcome :)

Also- don't know why I can't get my equations to format properly? ^^
< Moderator's note: fixed. You had a ( instead of { >

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Degeneracy of an operator's eigenvalue spectrum is usually a result of some symmetry. In your example the operator ##\frac{d^2}{dx^2}## is translation invariant which means that the set of all displaced sine functions of same wavelength, ##sin(a x + b)##, is a set of mutually degenerate states (a is kept constant and b can have any real value).

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hilbert2 said:
Degeneracy of an operator's eigenvalue spectrum is usually a result of some symmetry. In your example the operator ##\frac{d^2}{dx^2}## is translation invariant which means that the set of all displaced sine functions of same wavelength, ##sin(a x + b)##, is a set of mutually degenerate states.

Ahh okay that makes sense in this case. In general, do you know how this can be formally determined for any given operator (other than looking for a symmetry)?

Thanks :)

^ There are probably some methods to systematically find all the symmetries of a differential operator, but that's quite advanced group theory. There are also so called "accidental" degeneracies that aren't caused by symmetries, but there's probably no efficient way to find them if they exist.

vanhees71
Physgeek64 said:
if Q is a differential (for example ##\frac{p^2}{2m}= - \bar{h} \frac{d^2}{dx^2}## ) how can one write such an equation.

As a differential equation. For example, in your case we would have, for a function ##f(x)##,

$$- \hbar^2 \frac{d^2}{dx^2} f(x) - \lambda f(x) = 0$$

which obviously has solutions for ##f(x)## that are sines and cosines. This still doesn't guarantee finding a solution, since solving differential equations is not a completely mechanical process. But that's how you would start.

## 1. What does it mean for an operator to be degenerate?

Degeneracy in an operator refers to the existence of multiple eigenvalues with the same corresponding eigenvectors. This means that the operator has multiple solutions for a given input, making it more complex compared to a non-degenerate operator.

## 2. How can we determine if an operator is degenerate?

To determine if an operator is degenerate, we can find its eigenvalues and eigenvectors. If there are multiple eigenvalues that have the same eigenvector, then the operator is degenerate. We can also use the degeneracy theorem, which states that if an operator has a repeated eigenvalue, then it is degenerate.

## 3. What are the implications of an operator being degenerate?

Degeneracy in an operator can have significant implications in quantum mechanics and other areas of physics. It can affect the accuracy of measurements and predictions, and make the system more difficult to analyze. It also indicates that the system has more symmetries, which can have both positive and negative effects.

## 4. Can an operator be partially degenerate?

Yes, an operator can be partially degenerate. This means that some of its eigenvalues have corresponding eigenvectors, while others do not. In this case, the operator is partially complex and partially simple, making it easier to analyze compared to a fully degenerate operator.

## 5. How can we deal with degeneracy in an operator?

There are various methods for dealing with degeneracy in an operator, depending on the specific problem and application. In some cases, we can use perturbation theory to approximate the eigenvalues and eigenvectors of a degenerate operator. Other methods include degenerate perturbation theory, variational methods, and group theory. Ultimately, the approach will depend on the complexity of the operator and the desired level of accuracy.

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