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Given a free particle, the ground state of the system is the eigenstate with zero momentum. However, we also know that momentum operator does not have proper eigenstates, but rather it's spectrum is described in terms of projection-valued measure over the real axis. I am trying to understand how these two things can live together from a mathematical point of view. Any idea?

MBlrd
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  • A professor told me once that you can make sense of it mathematically, but we needed more advanced math to understand it, or make it consistent. May be he was bullsh... me. I see it as an analogy of needing generalized functions (or distributions) to understand some problems when standard calculus functions would not be up to the task. – Pato Galmarini Oct 19 '23 at 21:37
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    Put it in a box. – hft Oct 19 '23 at 21:57
  • @hft I am not sure that qualifies as a free particle, it is rather an infinite potential well, right? – Pato Galmarini Oct 19 '23 at 22:03
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    How else are you going to normalize the free particle wavefunctions? – hft Oct 19 '23 at 22:07
  • I mean, you can use wave packets, but those are not momentum eigenstates. I guess neither are particle in a box wavefunctions, but make the box very very big and how could the boundaries matter? – hft Oct 19 '23 at 22:08
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    Physicists play fast and loose with the math a little. A momentum "eigenstate" is a state $\lvert p \rangle$ of well-defined momentum $p$ whose position-space representation is $e^{ipx/\hbar}$. Mathematically, this isn't really an eigenstate of the momentum operator in the strict sense, and it's not normalizable, etc. The point is, the answer to your question is, "Physicists play fast and loose with the math and math terminology, but it's okay." – march Oct 19 '23 at 23:08
  • Normalizing the solution to free particle Schrödinger equation – mmesser314 Oct 20 '23 at 00:29
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    The idea of things like $|p\rangle$ can be mad rigorous in the framework of so-called Rigged Hilbert spaces. – Tobias Fünke Oct 20 '23 at 00:57
  • The momentum eigenstates are not single particle states. They serve as a basis in terms of which we can define single particle states. Physically, you can never have a particle with a perfectly precise momentum. – flippiefanus Oct 20 '23 at 03:12
  • Dirac uses that state (the "standard ket") in his book to base the definitions of all other states; it is the translationally invariant vacuum. – Cosmas Zachos Oct 20 '23 at 14:26
  • Linked. – Cosmas Zachos Oct 20 '23 at 16:42
  • Thank you @CosmasZachos! Are you saying that eigenstates of momentum don't exist (no "eigenket") but exist as functionals to represent every vector of the Hilbert space (exist as "eigenbra"). Is that right? Wouldn't this mean that the ground state of a free particle does not exist? – MBlrd Oct 21 '23 at 08:23
  • They exist suitably defined. Read up. – Cosmas Zachos Oct 21 '23 at 09:21
  • Quoting: "Given S⊂H, you can construct the space S∗⊃H of continuous (wrt the nuclear topology) complex-linear linear functionals on S"..."This space should be thought of as the space of bras, in the Dirac bra-ket sense. The bra ⟨x| is the linear function which maps ψ∈S to ψ(x)=⟨x|ψ⟩... (The space of kets is the conjugate space, consisting of conjugate-linear functionals on S. The ket |x⟩ maps a state ψ∈S to ψ∗(x)=⟨ψ|x⟩.)..."..."The elements of the algebra A can't have eigenvectors in H if they have continuous spectrum. But they do have eigenvectors in the space of bras." – MBlrd Oct 21 '23 at 11:32
  • That is what I meant before, in case of misunderstanding. I think what you are is saying matches what @DanielC is saying. Is my understainding correct? – MBlrd Oct 21 '23 at 11:34

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Actually, the question is derived from a misunderstanding which appears right in the first sentence. The ground state of a quantum system is defined as the (possibly generalized) eigenstate of the Hamiltonian (and not of the momentum operator) for its smallest spectral value, whereas the Hamiltonian by definition is bounded from below (*). The Hamiltonian of a free particle is a positive definite linear self-adjoint operator and its spectrum (which coincides with the spectrum of the tight-rigging extension of it) has the lower bound = $0$. So the PVM from von Neumann's spectral decomposition are built over $[0,\infty)$ (and not the full real line), which is the spectrum of the free-particle's Hamiltonian.

(*) A technical result is here: https://math.stackexchange.com/questions/4124505/does-the-spectrum-of-a-bounded-from-below-self-adjoint-operator-have-a-lower-bou

DanielC
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  • Hi @DanielC! Thank you for you answer, I am trying to digest it. Just to understand: you are saying that the ground state (i.e. the eigenstate of the hamiltonian with energy exactly equal to zero) exists since the eigenvalue = 0 is an isolated point of the spectrum. Is my understanding right? – MBlrd Oct 20 '23 at 19:41
  • Not isolated. If an operator is not bounded from below, it can't be a Hamiltonian. In other words, we must postulate (at least in standard Quantum Mechanics) the existence of a ground state. – DanielC Oct 20 '23 at 21:07
  • Sorry, I am not understanding how you are proving that, since the free hamiltonian is bounded then the eigenstate with energy equal to zero exists. If E=0 is not an isolated point in the spectrum, it does not have a proper eigenstate. Or not? – MBlrd Oct 20 '23 at 21:19
  • By this https://math.stackexchange.com/q/553013/, an isolated point in the spectrum is an eigenvalue, but $H$ has no isolated points. It has a continuous spectrum, hence any argument using isolated points doesn't apply. – DanielC Oct 21 '23 at 10:49
  • Ok, so you are saying that there is no "proper" ground state for a free particle. That is interesting, is this good from a purely physical point of view? – MBlrd Oct 21 '23 at 11:28