First, let me preface this statement by saying I know that there exists no (unique) self adjoint extension of the standard differential operator for the space $L_2([0,1])$.

However, when one attempts to actually prove this fact first they can go down the path of first enforcing $\psi(0) = \psi(1)$=0 as a boundary for the domain of $P$.

This actually makes the operator hermitian. However with this condition it is easily shown the adjoint of $P$ has a larger domain (namely no requirement of boundary conditions on $\psi$). Thus this operator as defined with its boundary condition can not be self adjoint.

But why can’t I simply use the operator $P = -i \frac{d}{dx}$ without any restriction on the domain? In other words, what the operator would be in the case of the real line. I understand that this wouldn’t be self adjoint, but isn’t the reason we want the operator to be self adjoint in the first place is so it admits a complete eigenbasis? However, the eigenstates of the real line operator surely span this smaller subspace. And isn’t this all we really need (that, and the eigenvalues being real which they are?).

Basically it seems we can span the whole space with generalized eigenfunctions ($\exp{(ikx)}$) using a moment operator $P$ that isn’t self adjoint. So why is this “wrong” to do?

  • $\begingroup$ Possible duplicate of physics.stackexchange.com/questions/446058/… $\endgroup$
    – DanielC
    Commented Jun 15, 2019 at 18:18
  • $\begingroup$ Daniel, I have seen that thread. However that is concerning the differences between P and P adjoint on the real line. This question is specifically asking why I can’t use P (without boundary conditions) as an operator whose eigenfunctions span the space (although it is not self adjoint) $\endgroup$ Commented Jun 15, 2019 at 18:23
  • $\begingroup$ Are you sure the periodic domain doesnt lead to welf adjoint operator? I was always under the impression if the exponential of the operator is the shift operator for x (which only works for periodic functions) the operator is self adjoint (alluding to stone von neumann). Of course the periodicity condition is more subtle, since L2 are equivalence classes, and points are a lebesgue zero set. $\endgroup$
    – lalala
    Commented Jun 15, 2019 at 19:22
  • $\begingroup$ @lalala, I actually edited it to mean the original condition of PSI(0) = 0 = PSI(1). But there are some boundary conditions indeed where it will be self adjoint. $\endgroup$ Commented Jun 15, 2019 at 19:52
  • $\begingroup$ Then exp(ikx) are not elements of your space (never zero) $\endgroup$
    – lalala
    Commented Jun 15, 2019 at 20:05

2 Answers 2


Taking the domain to be all functions $\phi(x)\in L^2[0,1]$ such that $-i\partial_x\phi \in L^2[0,1]$ certainly defines an opertor with $e^{ikx}$ as normalizable eigenfunctions, and the eigenfunctions span the space. These functions are not not mutually orthogonal though, and not even linearly indpendent in some $L^2$ sense. By this I mean that if we expand a function in $L^2[\mathbb R]$ whose support lies outside $[0,1]$ and the sum will converge to $0$. Consequently the functions span $L^2[0,1]$, but the expansions will not be unique, and therefore, I think, not vey useful.

  • $\begingroup$ Yes, I actually came to this conclusion myself as well before I saw this comment. But thank you! It does make sense. So in fact we can have operators that are not self adjoint that span the space, but in order to have a unique decomposition with a mutually orthogonal basis (which we do need in order to have a useful "observable" in the context of QM) then WE NEED self-adjoint operators. Cheers. $\endgroup$ Commented Jun 21, 2019 at 2:12

In the Schrödinger picture, for a "motion" in one dimension, the Hilbert space is typically $$ L^2 (\Omega), ~ \Omega\subseteq\mathbb R.$$ The exact value of this set (interval) $\Omega$ is specified by the dynamical conditions of the problem. This $\Omega$ is linked to the maximal range of the spectrum of the coordinate observable $x$. You have started with $ L^2 ([0,1])$, or more generally $ L^2 ([a,b])$, which tells you that the dynamics restricts the "motion" into a closed interval, which means the quantum particle is trapped, as in the case called "infinite potential well".

So the dynamical conditions fix the Hilbert space. Once you have this Hilbert space, you can try to define self-adjoint operators (which are called quantum observables). If this is possible, everything is ok. If not, then at least the Hamiltonian should be well-defined, in order to define the (Schrödinger) time evolution of states.

Now this $\Omega$ could very well be the whole $\mathbb R$, case in which the momentum operator is self-adjoint on its maximal domain of definition, $D(p)$.

But why can’t I simply use the operator P = -i d/dx without any restriction on the domain?

The domain $\Omega$ is specified by the dynamics, as I said above. For example, a harmonic oscillator can be defined by the condition: the "motion" is unrestricted on the real line, then the Hilbert space is $L^2 (\mathbb R)$, or the "motion" is restricted on the half line, then the Hilbert space is $L^2 [0,\infty)$, or the "motion" is restricted to the closed interval $[0,1]$. For each Hilbert space, you can check if the Hamiltonian, the coordinate or the momentum are properly defined and self-adjoint.

Regarding the boundary conditions, they are, too, a consequence of the dynamical restrictions. If the particle is trapped in potential well (let us choose the same interval $[0,1]$), then the boundary conditions $\psi (0) = \psi (1) = 0$ on the wave-functions are called physical, or natural, because the operators are differential operators, so it is expected a continuity of the wavefunction at the boundary. Outside of this box it is zero. This condition is not related to the symmetric or self-adjoint character of the momentum, it is given beforehand. Symmetry or self-adjointness are a consequence of this boundary condition, which is enforced on the wave-function in the domain of the operators.

  • 1
    $\begingroup$ Daniel, yes I understand the motion is restricted to L(0,1). What I was actually trying to understand is why I have to enforce some boundary conditions like PSI(0) = PSI(1) = 0 on P. I know that the point of enforcing these conditions is to try to force the operator to be self adjoint. However, isn’t the operator P with domain D(P) any weak differentiable function on L(0,1) able to span this whole space with the Fourier transform of EXP(ikx)? I know this is usually applied to the whole real line, but what’s stopping me just ignoring everything outside? This seems to span the space. $\endgroup$ Commented Jun 15, 2019 at 19:56
  • $\begingroup$ Added a few remarks. $\endgroup$
    – DanielC
    Commented Jun 16, 2019 at 18:51
  • 1
    $\begingroup$ But, it still seems to be that the Hilbert space we are working on is L2([0,1]) and not for example L2([0,1]) PSI(0) = PSI(L) = 0. We are applying the boundary conditions as a restriction of the domain of the operators. Yes, the operators are differential operators so their domain must be restricted s.t. they are continuous etc. But the space we are working in still allows for nonzero at the bounds. In other words, it really seems it is not given beforehand. Similar: link $\endgroup$ Commented Jun 19, 2019 at 4:04

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