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I know how to find the spectrum of the Hamiltonian to get the allowed energies for a system. If the spectrum is quantized, I can get definite values for each energy level. But when the system is in one of the corresponding eigenstates, does it actually have exactly that value? How does Heisenberg uncertainty come into play here?

Frank
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  • It doesn't - the uncertainty of an eigenstate of an observable with respect to that observable is zero. – ACuriousMind Mar 04 '15 at 18:27
  • So in the case of an eigenstate, the energy is very definite - but I was just looking at free particles in Griffiths, 2nd ed, and on page 60, he says free particles never have a definite energy... I am obviously confused about something here... – Frank Mar 04 '15 at 18:31
  • Who said that these free particles are in an eigenstate of the Hamiltonian? – ACuriousMind Mar 04 '15 at 18:33
  • Free particle states (in the terminology used by OP, I believe) are not normalizable and are therefore not in the Hilbert space. We always consider normalizable distributions of such states, which are not eigenstates of the Hamiltonian. – Prahar Mar 04 '15 at 18:35
  • OK - so they are in a superposition of the stationary eigenstates always, and therefore Heisenberg uncertainty kicks in? I'm going to have a follow-up question about photon absorption by electrons - the photons are initially "free" particles as they travel, and a beam of photons has a distribution of energies. When absorbed by an electron, only those with "the right energy" for the transition are going to be absorbed, right? Is that a definite energy? – Frank Mar 04 '15 at 18:37

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Heisenberg uncertainty is, in its general form as Robertson-Schrödinger uncertainty, given by

$$ \sigma_A \sigma_B \geq \frac{1}{2} \langle [A,B] \rangle $$

for any two observables $A,B$. Now, in an eigenstate $\lvert a \rangle$ of $A$, $\sigma_A$ vanishes, but also $\langle [A,B]\rangle = \langle a \rvert(AB - BA) \lvert a \rangle = \langle a \rvert( a B - B a )\lvert a \rangle = a \langle a \rvert( B - B )\lvert a \rangle = 0$, so the uncertainty relation is always obeyed.

Since you comment about Griffith's statement that a free particle never has definite energy, look at this:

A free particle has the Hamiltonian

$$ H = \frac{p^2}{m}$$

which has eigenfunctions $f_p(x) = \mathrm{e}^{\mathrm{i}px}$ if we go to the position representation where $p = \partial_x$.

Now, $f_p(x)$ is not square-integrable, and hence not a physical state, since physical states are described by wavefunctions in $L^2(\mathbb{R})$. Therefore, the eigenfunctions of the Hamiltonian are not true eigenvectors inside the Hilbert space, but only lie in a larger space that contains objects that are not normalizable. This observation leads one naturally to consider the idea of a Gel'fand triple.

Therefore, no physical state is an eigenstate of this Hamiltonian, and free particles never have definite energy. This has nothing to do with the uncertainty principle, and everything to do with the form of the Hamiltonian.

Phoenix87
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ACuriousMind
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  • I get it - thanks! The way I figured out it was not a physical state is by seeing that $\langle f_p|\hat{H}|f_p\rangle$ is undefined (the denominator blows up because $f_p$ is not square integrable - so to have a physically realizable $E$, you need to verify that the corresponding eigenstate is normalizable, not just stop at finding the spectrum of $\hat{H}$, right? – Frank Mar 04 '15 at 21:08
  • Minor comment but in general physical states are in $L^2(\mathbb{R}^3)$ :-) But it doesn't change anything to this discussion to stay in 1D. – Frank Mar 04 '15 at 21:12
  • @Frank: Fully correct. As a general rule: If the spectrum is continuous, i.e. if you find eigenvalues $E+\epsilon$ for every $\epsilon\in[0,e)$ for some $E,e$, it is not possible that the corresponding states are physical states, since they would have to be part of an orthonormal basis, but the basis of the space of physical states $L^2(\mathbb{R}$ or $L^2(\mathbb{R}^3)$ or any other separable space is countable, but every interval in $\mathbb{R}$ is uncountable. – ACuriousMind Mar 04 '15 at 21:15
  • Amazing! So "continuous or semi-continuous spectrum" means "there are non physical states"! Weird though, because "free" systems tend to have continuous spectra (no boundary condition to impose quantization) - so those "free" systems $H$'s eigenvalues/vectors are never going to directly be possible energies/states of the system... – Frank Mar 04 '15 at 21:21
  • I am not sure I understand your comment about basis though - since for e.g., the particle in a well, we can have more than 3 eigenfunctions that would all together form a basis in which to write the wavefunction - shouldn't the basis be a basis of a function space? – Frank Mar 04 '15 at 21:25
  • @Frank: Have a look at my answer here. The basis of the function spaces $L^2$ is countable, i.e. not finite, but it is strictly smaller than the size of $\mathbb{R}$. – ACuriousMind Mar 04 '15 at 21:30
  • I take back my question - you are absolutely right of course, I wasn't paying attention for a second :-) – Frank Mar 04 '15 at 21:32
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I believe that the relation between admissible values of an observable and Heisenberg uncertainty principle is best seen in the axiomatic formulation of quantum mechanics.

Starting from the postulate that every quantum mechanical system is described by a C*-algebra $A$, an admissible value $\lambda$ for the observable $O\in A$ is the value of a state $\omega$ from the state space of $A$ evaluated over $O$, that is $$\lambda = \omega(O),$$ with the extra condition that $$\Delta_\omega(O) := \sqrt{\omega(O^2-\omega(O)^2I)} = 0.$$

Now $\lambda$ belongs to the so-called physical spectrum of the observable $O$, but it can be shown through the GNS construction that this is indeed the algebraic spectrum of $O$ when viewed as an element of the C*-algebra $A$. To see this, do the GNS construction associated to $\omega$. You get a cyclic vector $\xi$ and a representation $\pi$ such that $\xi$ is a $\lambda$-eigenvector for $\pi(O)$. Therefore $\pi(O)-\lambda I$ is not invertible, which implies that $O-\lambda 1_A$ is not invertible in $A$, i.e. $\lambda\in\sigma(O)$, where $\sigma(O)$ is the algebraic spectrum of $O$ w.r.t. $A$.

Phoenix87
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Heisenberg Uncertainty Principle is always applicable. The energy eigenstate will remain so for an infinite time as long as a perturbation leads it to decay.

SAKhan
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