In a lecture on Quantum mechanics, the professor concluded that if $a$ is a linear operator with $[a, a^\dagger] = 1$, where $a^\dagger$ is the adjoint of $a$ and $[a, a^\dagger] = aa^\dagger - a^\dagger a$, and if $a^\dagger a$ has an eigenvalue, then vectors of form \[ | n \rangle = (a^\dagger)^n | 0 \rangle \quad ( n \in \mathbb Z_{ \ge 0 }) \] are all eigenvectors of $a^\dagger a$. ($| \lambda \rangle$ is the eigenvector of $a^\dagger a$ which belongs to the eigenvalue $\lambda$).

But he gives no reason why its eigenvalue exists. So, I asked him if an eigenvalue exists, and he said 'I cannot answer. Roughly speaking, it is some limit of the case of a finite dimensional complex vector space.'

My question is: Let $H$ be an infinite dimensional Hilbert space, say $\ell^2$ space, whose inner product is denoted by $\langle \cdot, \cdot \rangle$. It is well-known that for a bounded linear operator $T \in \mathcal L(H)$, there corresponds a (bounded) linear operator $T^\dagger \in \mathcal L(H)$ such that $\langle x, Ty \rangle = \langle T^\dagger x, y \rangle$ for all $x, y \in H$. Then, is it true that for any bounded linear operator $T \in \mathcal L(H)$ with $[T, T^\dagger] = 1$, there exists a scalar $\lambda \in \mathbb C$ and a nonzero vector $x \in H$ such that $T^\dagger T x = \lambda x$?

How could I deal with it? Is there a 'characteristic polynomial' of $T^\dagger T$?

Please give me the answer, or some references.


closed as off-topic by Yemon Choi, Pedro Lauridsen Ribeiro, David Handelman, Andreas Blass, Christian Remling Jan 21 '18 at 18:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Yemon Choi, Pedro Lauridsen Ribeiro, David Handelman, Andreas Blass, Christian Remling
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 17
    $\begingroup$ There are no bounded linear operators $A,B$ satisfying $[A,B]=1$. I think your professor was not making precise mathematical statements $\endgroup$ – Yemon Choi Jan 20 '18 at 18:49
  • 5
    $\begingroup$ The result that Yemon Choi quoted is the Wintner-Wielandt theorem: de.wikipedia.org/wiki/Satz_von_Wintner-Wielandt $\endgroup$ – GH from MO Jan 20 '18 at 19:36
  • 3
    $\begingroup$ (With unbounded operators it is possible, and the example is: A= d/dx and B=multiplication by x). $\endgroup$ – Pietro Majer Jan 20 '18 at 22:59
  • 1
    $\begingroup$ The books of Reed and Simon are good reference works on basic linear functional analysis. $\endgroup$ – Ben McKay Jan 21 '18 at 9:35

As noted in comments, there are no bounded operators whose commutator with their adjoint is a non-zero scalar. The prototype of unbounded operators with the desired property is the 1-dimensional Dirac operator $T=ix\pm i{d\over dx}$ on $L^2(\mathbb R)$. [Correction, thanks to @ChristianRemling for noticing my blunder!) The composition $T\circ T^*$ (not commutator, jeez) of it with its adjoint is (up to sign) $-\Delta+x^2+1$, essentially the quantum harmonic oscillator. The commutator is a constant (specifically, $\pm2$, the way I've written it. Whew...) Because the "confining potential" is present, this has compact resolvent, hence, purely discrete spectrum, and all the eigenvalues are indeed obtained as indicated.

In fact, the Stone-vonNeumann theorem essentially proves the uniqueness of such a "model" for the commutation relations, by proving that there is a unique irreducible repn of a Heisenberg group having a given central character. That is, up to some sort of isomorphism, the configuration of Hilbert space, $T$, and $T^*$ is uniquely determined. Thus, approximately, every such $T,T^*$ behave in the same way, so should have pure point spectrum, etc.


Not the answer you're looking for? Browse other questions tagged or ask your own question.