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34 votes
Accepted

What's the deal with momentum in the infinite square well?

Is the momentum operator $P=-i\hbar \frac{\mathrm d}{\mathrm dx}$ symmetric when restricted to the compact interval of the well? Are there any subtleties in its definition, via its domain or similar, ...
Valter Moretti's user avatar
31 votes
Accepted

Why do eigenvalues correspond to observable quantities?

Suppose we don't know quantum mechanics yet and we want to calculate the expecatation value of an observable $A$. Could be momentum, spin whatever. It is given by $$\mathbb E(A)=\sum_i a_i\,p(a_i)$$ ...
AccidentalTaylorExpansion's user avatar
30 votes
Accepted

Why hermitian, after all?

Hermitian operators (or more correctly in the infinite dimensional case, self-adjoint operators) are used not because measurements must use real numbers, but rather because we almost always decide to ...
Physics Footnotes's user avatar
27 votes

Must observables be Hermitian only because we want real eigenvalues, or is more to that?

At the simplest level it is because Hermitian, or more precisely self adjoint, operators have a complete set of eigenstates. The existence of a complete set is essential for the propability ...
mike stone's user avatar
  • 54.5k
27 votes

Why do we observe particles, not quantum fields?

We don't observe particles, at least not in the sense of the physical definition of a particle (as the physical approximation of the motion of an extended classical body by the motion of its center of ...
FlatterMann's user avatar
  • 3,075
26 votes
Accepted

Is hermiticity a basis-dependent concept?

The relation $$ \langle Ay | x \rangle = \langle y | A x \rangle \text{ for all } x \in \text{Domain of } A\tag1$$ makes no reference to any basis at all, so it is indeed basis-independent. In fact,...
Emilio Pisanty's user avatar
25 votes

Heisenberg uncertainty principle in daily life

If by "daily life" you mean things we experience on a day to day basis at the macroscopic level (as opposed to the microscopic level), while the principle applies it does so insignificantly. The ...
Bob D's user avatar
  • 73.7k
24 votes
Accepted

Not all self-adjoint operators are observables?

The appendix of Mackey talks about superselection rules, and indeed superselection is the phenomenon where there are self-adjoint operators that are not observables. Whether this is obvious or not ...
ACuriousMind's user avatar
  • 126k
24 votes
Accepted

Are eigenstates of the position operator continuous?

This is a good question, and the answer depends on how mathematically watertight we want to be. In the standard physicist's presentation of quantum mechanics, we start with a Hilbert space $\mathcal H$...
J. Murray's user avatar
  • 70.2k
24 votes
Accepted

Why does the expectation value in quantum mechanics correspond to the classically measured value?

In general, there is no such thing as a "classically measured position" for a generic quantum system/state. Some situations are simply not well-modeled by classical physics, and Ehrenfest's ...
ACuriousMind's user avatar
  • 126k
23 votes

Not all self-adjoint operators are observables?

In general, for physical reasons not all self-adjoint operators are observables. I will discuss the problem at the level of von Neumann algebras that are (perhaps unawares) more familiar to physicists....
Valter Moretti's user avatar
23 votes

Why do we choose the Dirac delta function as the eigenstate of position operator?

In elementary treatments, wavefunctions are sometimes described as $\mathbb C$-valued functions which are square-integrable, i.e. $$\int_\mathbb R |\psi(x)|^2 \mathrm dx < \infty$$ This is almost ...
J. Murray's user avatar
  • 70.2k
23 votes
Accepted

Why does the Heisenberg uncertainty principle apply to particles?

I understand your confusion. It is due to an old-fashioned way of introducing Uncertainty relations based on wave formalism, dating back to Heisenberg, but probably quite misleading. Quantum mechanics ...
GiorgioP-DoomsdayClockIsAt-90's user avatar
23 votes

Definition of four-velocity: why define it with proper time of the object?

$dx^\mu$ is covariant and $d\tau$ is invariant. So $dx^\mu/d\tau$ is manifestly covariant while $dx^\mu/dt$ is not. Covariant quantities are of interest because the laws of physics are covariant. So ...
Dale's user avatar
  • 103k
22 votes

Heisenberg uncertainty principle in daily life

While I agree with tparker saying "If simple examples of the Heisenberg uncertainty principle occurred in everyday life, then it would have been discovered before 1927.", your question reminded me of ...
Stipe Galić's user avatar
22 votes
Accepted

Why are expectation values of an observable important in QM?

This is an important point which should be discussed. What one obtains from experiments are frequencies of outcomes of given measured observables on an ensemble of identical quantum systems all ...
Valter Moretti's user avatar
21 votes

In what sense (if any) is Action a physical observable?

The book propagates a myth. Experiments measure angular momentum, not action - even though these have the same units. One finds empirically that angular momentum in any particular (unit length) ...
Arnold Neumaier's user avatar
21 votes
Accepted

Why do we use Eigenvalues to represent Observed Values in Quantum Mechanics?

One of the postulates of quantum mechanics is that for every observable A, there corresponds a linear Hermitian operator A^, and when we measure the observable A, we get an eigenvalue of A^ as the ...
J. Murray's user avatar
  • 70.2k
20 votes

Are observables in QFT actually observable?

A quantum mechanical secret kept very well in plain sight is that the word "observable" when defined as "self-adjoint operator on a Hilbert space" does not actually mean "you ...
ACuriousMind's user avatar
  • 126k
19 votes

Do we or do we not observe (measure) superpositions all the time?

To answer the question, you have to specify what observable’s eigenstates the “superpositions” you’re talking about are superpositions of. If you measure the energy, you always get one single energy ...
G. Smith's user avatar
  • 51.7k
19 votes

Really why does promoting numerical variables to operators neatly work?

...why does promoting numerical variables to operators neatly work? The question seems to be asking about quantization, a recipe for constructing a quantum model based on a given classical model. Why/...
Chiral Anomaly's user avatar
17 votes

Heisenberg uncertainty principle in daily life

Maybe I missed it, but somehow no-one seems to have mentioned the Bandwidth theorem. The Heisenberg Uncertainty Principle is simply one manifestation of it. Basically if you have a function $\psi\left(...
Cryo's user avatar
  • 3,391
17 votes
Accepted

Are two states with the same measurement probabilities necessarily equal up to unitary equivalence?

Let $H$ denote a finite-dimensional complex Hilbert space, $\rho$ and $\rho^\prime$ be two density matrices, i.e. positive semi-definite operators with unit trace and $A$ an arbitrary hermitian ...
Tobias Fünke's user avatar
17 votes

How to interpret quantum fields?

There is one usual confusion about quantum fields which is, at least in my perspective, perhaps caused by one very familiar example which we all have met before studying QFT. This example is that of ...
Gold's user avatar
  • 36.4k
16 votes

If I was Schrödinger's cat, what would I feel?

In practice, you can't actually get yourself into a quantum superposition of dead and alive. Exposing a quantum system to a thermal environment causes the state to lose coherence and collapse into one ...
Chris's user avatar
  • 17.2k
16 votes

Do states with infinite average energy make sense?

If you adopt the frequentist perspective, the fact that $\langle E\rangle$ does not exist simply means that if you measure the energies of $N$ identically-prepared systems and average the results, ...
J. Murray's user avatar
  • 70.2k
16 votes

Why do we observe particles, not quantum fields?

"My understanding is that particles arise as a computational tool." This is a "cart before the horse" statement. Experiments determined that particles existed,( see here a bubble ...
anna v's user avatar
  • 234k
15 votes
Accepted

Does a complete set of commuting observables always exist?

Sometimes textbooks on introductory quantum theory give the impression that every self-adjoint operator represents an observable, and many simple models do have that property. Such simple models ...
Chiral Anomaly's user avatar
15 votes
Accepted

Is a quantity calculated from observables, observable?

Is a quantity calculated from observables, observable? A quantity calculated only from observables is observable. Example 1) If we view the mass and velocity of a classical particle as observables, ...
Dale's user avatar
  • 103k
14 votes
Accepted

Is every observable a function of position and momentum?

In classical physics, the definition of an observable is that it is a (reasonably smooth) function of position and momentum. Therefore, quantum systems obtained purely by quantization of a classical ...
ACuriousMind's user avatar
  • 126k

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