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Without reproducing proofs:

  1. Eigenvalues of a Hermitian operator are real (proof does not rely on the boundary conditions).

  2. The momentum operator is Hermitian (proof does not rely on the boundary conditions).

  3. Without any boundary conditions, eigenvalues of the momentum operator can be complex.

How is this possible? Proofs of 1, 2, and 3 can be found in most introductory texts, but I can reproduce them if their veracity is in question.

Added: By Hermitian I mean $\int f^* (D g) \, d^3r=\int (Df)^* g\, d^3r$. The proof of 2 rely on $fg\to 0$ as $x\to\infty$. The boundary conditions that force eigenvalues of the momentum operator to be real is the periodicity of the eigenfunction.

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    $\begingroup$ The boundary conditions determine whether an operator is Hermitian or not. Once you know your operator is Hermitian, you have the results on the spectrum. Without boundary conditions the momentum operator need not be Hermitian, hence its spectrum can have non-real values (here I am assuming that by Hermitian you actually mean self-adjoint). $\endgroup$ – Phoenix87 Jul 18 '16 at 13:10
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    $\begingroup$ @Phoenix87 please post that as an answer at your earliest convenience. The fact that boundary conditions are an essential part of what gives differential operators their identity is too often ignored, resulting in much confusion. $\endgroup$ – DanielSank Jul 18 '16 at 13:15
  • $\begingroup$ Your point 2 is flat-out incorrect. Removing boundary conditions makes $⟨x|\psi⟩=e^x$ a valid wavefunction, for which $⟨\phi|\hat p|\psi⟩$ is completely undefined (taking, say, $⟨x|\phi⟩=1/(1+x^2)$), let alone something that you can actually use in a proof. $\endgroup$ – Emilio Pisanty Jul 18 '16 at 13:50
  • $\begingroup$ @Phoenix87 Every symmetric operator has real numerical range (where the numerical range is the set of values $\langle u,Tu\rangle$, $u\in D(T)$ with $\lVert u\rVert=1$), as it follows easily from the definition. Therefore symmetric operators in general are such that the eigenvalues, if there are any, must be real. $\endgroup$ – yuggib Jul 18 '16 at 13:51
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The boundary conditions determine whether an operator is Hermitian or not. Once you know your operator is Hermitian, you have the results on the spectrum. Without boundary conditions the momentum operator need not be Hermitian, hence its spectrum can have non-real values (here I am assuming that by Hermitian you actually mean self-adjoint).

As an example, consider the operator $$D:=i\frac{\text d}{\text ds}$$ acting on the Hilbert space $L^2([0,1])$. For its domain we can choose the proper subspace $H_0^1$ of the Sobolev space $H^1$ containing all those functions that vanish on the boundary of the closed interval. The adjoint $D^*$ of $D$ can be defined on the whole of $H^1$, hence $D$ is Hermitian, but not self-adjoint. Indeed the defects are $$n_\pm(D)=1$$ and there is a one-parameter family of self-adjoint extensions of $D$, parametrised by the unit circle $S^1$.

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I will answer with an example on how eigen functions of momentum do not exist in a Hilbert space in general. When they do not, we say the momentum operator is not self-adjoint.

Consider a one dimensional infinite potential well. Lets us place the walls at $x=0$ and $x=L$. For $x\ge L$ and $x\le 0$, the potential $V(x)=\infty$ and therefore we put the boundary condition for the wave function $\psi(0)=0=\psi(L)$.

In the Hilbert space of wavefunctions that satisfy this boundary condition, the momentum operator is not self adjoint. Let us see how.

The eigen value equation for momentum operator in the coordinate basis is $-i\frac{\partial\psi(x)}{\partial x}=p\psi(x)$. But this equation has no solution with the above boundary condition. i.e. $\psi(x)=e^{ipx}$ does not exist in this Hilbert space. Therefore momentum as an operator is said to be not self-adjoint under this boundary condition. The eigenfunctions of momentum simply do not exist in this space.

P.S. I realize now that I have merely put in simpler words through an example, what Pheonix87's answer says.

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