- #36
Antarres
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I'm not an expert on functional analysis, but I will try to solve this, since it looks like a good practice. Hopefully, I didn't miss the point of the exercise.
Eigenvectors of given operators will be real functions, for which we need to solve differential equations:
$$Af = \lambda f \qquad Bf = \lambda f \qquad Cf = \lambda f$$
Here ##\lambda## is an eigenvalue.
All of these equations are simple equations which are solved by separation of variables. Take the first one for example:
$$2x\frac{df}{dx} = \lambda f(x)$$
Separate:
$$2\frac{df}{f(x)} = \lambda\frac{dx}{x}$$
Integrating both sides, after short and simple algebra, we find a class of functions which are eigenvectors to the operator ##A##:
$$f(x) = Cx^{\frac{\lambda}{2}}$$
Where ##C## is arbitrary real number(integration constant).
We similarly solve for eigenvectors of ##B## and ##C##:
$$Bf = \lambda f \Leftrightarrow x^2\frac{df}{dx} = \lambda f(x) \Rightarrow \frac{df}{f(x)} = \lambda \frac{dx}{x^2} \Rightarrow f(x) = Ce^{-\frac{\lambda}{x}}$$
$$Cf = \lambda f \Leftrightarrow -\frac{df}{dx} =\lambda f(x) \Rightarrow \frac{df}{f(x)} = -\lambda dx \Rightarrow f(x) = Ce^{-\lambda x}$$
where the last implication involves integrating both sides and rearranging terms in appropriate way, and ##C## are arbitrary real numbers.
Now we inspect the multiplication structure of the given space. It's obvious that it's not closed under composition, since composition of any of those operators would involve second derivatives, but it looks like it is closed under commutators. So we will check for commutator algebra:
$$[A,B]f = 2x\frac{d}{dx}\left(x^2\frac{df}{dx}\right) - x^2\frac{d}{dx}\left(2x\frac{df}{dx}\right) = 2x^2\frac{df}{dx} = 2Bf$$
$$[B,C]f = x^2\frac{d}{dx}\left(-\frac{df}{dx}\right) +\frac{d}{dx}\left(x^2\frac{df}{dx}\right) = 2x\frac{df}{dx} = Af$$
$$[C,A]f = -\frac{d}{dx}\left(2x\frac{df}{dx}\right) -2x\frac{d}{dx}\left(-\frac{df}{dx}\right) = -2\frac{df}{dx} = 2Cf$$
So we have the following relations:
$$[A,B] = 2B \qquad [B,C] = A \qquad [C,A] = 2C$$
So the space spanned by these three operators is closed under commutator operation, so it forms a Lie algebra with structure constants given above(the only nontrivial ones).
Was this the whole object of the exercise? If I'm missing something, please point it out so I can add it. Thanks!
$$Af = \lambda f \qquad Bf = \lambda f \qquad Cf = \lambda f$$
Here ##\lambda## is an eigenvalue.
All of these equations are simple equations which are solved by separation of variables. Take the first one for example:
$$2x\frac{df}{dx} = \lambda f(x)$$
Separate:
$$2\frac{df}{f(x)} = \lambda\frac{dx}{x}$$
Integrating both sides, after short and simple algebra, we find a class of functions which are eigenvectors to the operator ##A##:
$$f(x) = Cx^{\frac{\lambda}{2}}$$
Where ##C## is arbitrary real number(integration constant).
We similarly solve for eigenvectors of ##B## and ##C##:
$$Bf = \lambda f \Leftrightarrow x^2\frac{df}{dx} = \lambda f(x) \Rightarrow \frac{df}{f(x)} = \lambda \frac{dx}{x^2} \Rightarrow f(x) = Ce^{-\frac{\lambda}{x}}$$
$$Cf = \lambda f \Leftrightarrow -\frac{df}{dx} =\lambda f(x) \Rightarrow \frac{df}{f(x)} = -\lambda dx \Rightarrow f(x) = Ce^{-\lambda x}$$
where the last implication involves integrating both sides and rearranging terms in appropriate way, and ##C## are arbitrary real numbers.
Now we inspect the multiplication structure of the given space. It's obvious that it's not closed under composition, since composition of any of those operators would involve second derivatives, but it looks like it is closed under commutators. So we will check for commutator algebra:
$$[A,B]f = 2x\frac{d}{dx}\left(x^2\frac{df}{dx}\right) - x^2\frac{d}{dx}\left(2x\frac{df}{dx}\right) = 2x^2\frac{df}{dx} = 2Bf$$
$$[B,C]f = x^2\frac{d}{dx}\left(-\frac{df}{dx}\right) +\frac{d}{dx}\left(x^2\frac{df}{dx}\right) = 2x\frac{df}{dx} = Af$$
$$[C,A]f = -\frac{d}{dx}\left(2x\frac{df}{dx}\right) -2x\frac{d}{dx}\left(-\frac{df}{dx}\right) = -2\frac{df}{dx} = 2Cf$$
So we have the following relations:
$$[A,B] = 2B \qquad [B,C] = A \qquad [C,A] = 2C$$
So the space spanned by these three operators is closed under commutator operation, so it forms a Lie algebra with structure constants given above(the only nontrivial ones).
Was this the whole object of the exercise? If I'm missing something, please point it out so I can add it. Thanks!
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