I am a freshman trying to understand the very basics of quantum mechanics but I met barriers at the beginning. What really matters is the postulates of quantum mechanics and their relationship with self-adjointness.

Postulate 1) Every observable of a physical system is represented in the mathematical formalism of quantum mechanics by a linear adjoint operator which acts in the Hilbert space associated with the physical system considered.

As far as I know, self adjointness of a given form of operator is dependent on the format of Hilbert space. For example, momentum operator, represented by first-derivative, is not self adjoint in infinite well but in free space. Then for the given observable (here, momentum), the form of operator correspond to the observable should be changed to make the operator self adjoint when we use it in different kinds of Hilbert spaces??

Furthermore if someone wants quantum mechanics to be used with strict mathematical formalism, should he/she always check self-adjointness of the given operator everytime he/she uses it in different Hilbert spaces?

Postulate 2) If B is a Hermitian operator that represents physically observable property, then the eigenfunctions of B form a complete set for the Hilbert space considered.

If postulate 1 is true, shouldn't we change the word "Hermitian" used in postulate 2 with "self adjoint"??

And.... after we change the word, is it still correct that two commuting (self adjoint) operators share common eigenfunctions?

Postulate 3) The time dependence of the state of an undisturbed quantum mechanical system is given by Schrodinger equation..... and if we assume that the state is stationary, the form of the equation is Hf = Ef where f is the wave function, H is Hamiltonian, and E is energy of the system.

Because energy is a sort of observables, this postulate tells us that Hamiltonian operator is always self adjoint regardless of the choice of Hilbert spaces if postulate 1 is true??

And.. If we solve the (Schrodinger) equation and get a set of eigenfunctions (of Hamiltonian) for the system, do the eigenfunctions have all the information about the system? If the eigenfunctions are not the eigenfunctions of... say, p, which means that p don't commute with H, the eigenfunctions of p cannot be one of the possible states of the system?

  • $\begingroup$ Self-adjointes does not depend on the Hilbert space. I mean, changes of ``representations'' are performed by means of unitary operators $A \to UAU^{-1}$ and they do not change self-adjointness properties of operators $A$. There is no guarantee to have a complete set of eigenvectors (more precisely a spectral measure) if the operator is only Hermitian and not self-adjoint, so the true condition on observables is self-adjointness and not Hermiticity... $\endgroup$ – Valter Moretti Dec 4 '14 at 11:07
  • $\begingroup$ @Daan Sim: who told you that in an infinite well the linear momentum operator is not self-adjoint? And beware, this operator is NOT the derivative, but -ihbar multiplied by the first derivative. About checking each time, no it's not necessary. $\endgroup$ – Sofia Dec 4 '14 at 11:19
  • $\begingroup$ Sofia, the problem is that $p$ acting on states shifts them rightwards, out of the Hilbert space, so to speak. It's a matter of interpretation, because some people take the infinite well to be equivalent to a finite interval. For square integrable functions on that interval, $p$ is not self adjoint. $\endgroup$ – lionelbrits Dec 4 '14 at 16:11
  1. If you have different Hilbert spaces, you cannot say it is the same operator on them, since operators are defined on the Hilbert space. The momentum operator is a tricky one for many systems, and rigor requires the discussion of concepts like rigged Hilbert spaces. A nice introductory discussion of this is "Mathematical surprises and Dirac's formalism in quantum mechanics" by Francois Gieres.

  2. Yes. The spectral theorem holds for self-adjoint operators, not Hermitian ones. Those two notions only coincide on finite-dimensional Hilbert spaces, but physicists are (sadly) often sloppy about this because they don't want introductory quantum mechanics to turn into full-blown functional analysis.

  3. The Hamiltonian must be self-adjoint since energy is an observable, yes. It is possible to relax the demand of self-adjointness and demand only a PT-symmetric Hamiltonian, and one can still obtain a reasonable quantum theory, but this is rather exotic. In all usual contexts, the Hamiltonian is self-adjoint and observable. Since it is self-adjoint, its eigenvectors indeed span the whole space. But every vector is allowed to be a state of the system, and just because $p$ doesn't commute with $H$ doesn't mean eigenstates of $p$ are disallowed - how could they, given that, if you measure $p$, you will find the system in one of its eigenstates by assumption? The non-commutativity just means that you can never have a system simultaneously in an eigenstate of both non-commuting operators - if it is the eigenstate of one of them, it will be a linear combination of eigenstates of the other (since they form a basis!)-

  • $\begingroup$ The Hamiltonian has COMPLEX eigenvalues when one tries to obtain the wave-function that describes the decay (the Gamow state). But, for a BEGINNER in quantum theory, I can't get into such things. Before explaining to someone the integral calculus, he has to understand simple algebraic calculus. $\endgroup$ – Sofia Dec 4 '14 at 15:26
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    $\begingroup$ @Sofia: That is merely an artifact of trying to describe a situation with QM that QM isn't really made for - whenever particle number is not conserved, the natural description is in terms of a full quantum field theory, where the Hamiltionian is self-adjoint again. $\endgroup$ – ACuriousMind Dec 4 '14 at 15:44
  • $\begingroup$ @CuriousMind: It's very late in my country. I just noticed your reaction, and I am half-reacting. Of course QM has difficulty with the decay. But I reject your statement that in the decay the number of particles is not conserved. I currently deal with the NUCLEAR alpha decay. Before being emitted, the alpha is prepared inside the parent nucleus. And about the Hamiltonian, you know that if it is self-adjoint, the process is reversible. Well, the alpha emission IS NOT REVERSIBLE. But, it is very late, let's leave this argument for tomorrow. Good night! $\endgroup$ – Sofia Dec 5 '14 at 0:47

Ok, there are a lot of points here.


First of all, an operator in Hilbert spaces is not defined only by its action (e.g. the operation of derivation for the momentum), but also by the so-called domain of definition, i.e. the subspace of vectors of the Hilbert space where it can act. Unbounded operators are not defined for every vector of the Hilbert space but only on a dense subset called the domain of the operator. Given an operator (i.e. given its action and its domain), one can ask if the operator is closed w.r.t. that domain, self-adjoint, etc... Suppose you have a closable densely defined symmetric operator $A$. Then it may have zero, one or infinite self-adjoint extensions. The derivation operator--densely defined on the continuous infinitely differentiable functions--has only one self-adjoint extension in $L^2(\mathbb{R})$; the derivation operator defined on $\{\psi, \psi\in AC(0,1), \psi'\in L^2([0,1]), \psi(0)=0=\psi(1)\}$ has no self-adjoint extensions in $L^2([0,1])$, but if it is defined on $\{\psi, \psi\in AC(0,1), \psi'\in L^2([0,1]), \psi(0)=\alpha\psi(1)\}$, with $\lvert\alpha\rvert=1$ then it is self-adjoint.

So the problem is not that you need to change the form of an operator when you use it in different spaces, simply you have to consider operators as a whole, i.e. as an action and a corresponding domain of definition. Once you have done that, you may inquire whether it is closed, symmetric, self-adjoint, essentially self-adjoint, etc.


The only self-adjoint operators such that the corresponding eigenvectors form a basis of the Hilbert space are those that are compact, or with compact resolvent (self-adjointness and hermitcity are not the same thing for unbounded operators). And the most correct form of the other assertion on commutation is that two commuting self-adjoint operators share a common spectral family of projections (I cannot be more precise on this point with your level of knowlege). Let's say that if they are both compact or with compact resolvent then the assertion you have made is true.


The postulate should read: "The dynamics of the system is generated by a unitary group of transformations". These unitary groups are in one-to-one correspondence with self-adjoint operators (Stone's theorem). The latter act as generators. So the Hamiltonian---since it is exactly the object defined to be the generator of the dynamics---has to be self-adjoint in quantum mechanics.


A small addition on observables (your second postulate). It is tricky to define in a mathematically precise way quantum mechanical observables. A nice and elegant mathematical way is to define observables as a $C^*$-algebra, i.e. an algebra of self-adjoint bounded operators on an Hilbert space, and formulate axioms of quantum mechanics starting from observables rather than Hilbert spaces themselves. A drawback is that unbounded operators are not included as observables with this point of view, while a lot of quantities that are considered to be observables in physics are unbounded operators: energy and momentum above all. I suppose that a very minimal requirement for a physical observable in quantum mechanics is that it is a symmetric operator on a Hilbert space, so that its numerical range is a subset of the real line (i.e. its expectation value is always a real number).


Quantum mechanics is a physical theory. For a given physical system (set up and possible states) you can fix a Hilbert space, and some linear operators. Some linear operators will be unitary (for instance time evolution, to advance a state at one time to a state at a different time), some operators will be self-adjoint (for instance for an observable).

An linear operator is technically a linear map from (a subspace of) the Hilbert space to itself, so when you have a different Hilbert space you technically have different operators by definition.

For an observable, you do also generally want the eigenfunctions to be complete.

For two commuting observables, the eigenfunctions of one may or may not be eigenfunctions of the other, but there are common eigenfunctions, and enough of them.

Time dependence is given by the Schrödinger equation, and for stationary states Hf=Ef, where f is the wave function, H is Hamiltonian, and E is the energy of the system.

Sometimes you want more from an operator than merely that it be self-adjoint, for instance you might want it to have a finite expectation value for every state, because sometimes you consider the expectation value to itself be a kind of ensemble-observable.

Do the eigenfunctions of the Hamiltonian have all the information about the system?

The eigenfunctions of the Hamiltonian are complete, and you know how each evolves (because you know the energy), so can you figure out how any bounded operator acts at any time. That's pretty good. (For an unbounded operator, you'd have to check to see if your state is in the domain)

If the eigenfunctions are not the eigenfunctions of... say, p, which means that p doesn't commute with H, the eigenfunctions of p cannot be one of the possible states of the system?

If an observable does not commute with the Hamiltonian, then they won't have common eigenfunctions but that's totally fine. That just means that after a measurement of the momentum observable, it will not be an energy eigenstate even if it was one before. But the energy eigenfunctions are still complete, so you'll still know how that momentum eigenfunction evolves.

Which leads to the things I didn't see in your post, a single thing about measurement or projection.


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