57
votes
Is there something behind non-commuting observables?
Observables don't commute if they can't be simultaneously diagonalized, i.e. if they don't share an eigenvector basis. If you look at this condition the right way, the resulting uncertainty principle ...
- 101k
33
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, ...
- 70k
30
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
$$\mathbb E(A)=\sum_i a_i\,p(a_i)$$
...
28
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 ...
- 2,676
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,...
- 131k
26
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 ...
- 51.2k
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 ...
- 69k
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 ...
- 122k
23
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 $\mathcal H$...
- 67.5k
23
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 ...
- 122k
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 ...
- 67.5k
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 ...
- 94.5k
22
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....
- 70k
22
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 ...
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) ...
- 44.5k
21
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 ...
- 680
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 ...
- 67.5k
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 ...
- 122k
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 ...
- 51.3k
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/...
- 53.6k
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 ...
- 35.1k
17
votes
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 ...
- 70k
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 ...
- 17.1k
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, ...
- 67.5k
16
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 ...
- 7,175
15
votes
Is there a time operator in quantum mechanics?
Just open any string text which has a discussion of the relativistic point particle.
http://arxiv.org/abs/0908.0333 - Section 1 for example
or Green, Schwartz, Witten Volume 1
Punchlines:
1) Time ...
- 833
15
votes
Accepted
Heisenberg's uncertainty principle for mean deviation?
We can assume WLOG that $\bar x=\bar p=0$ and $\hbar =1$. We don't assume that the wave-functions are normalised.
Let
$$
\sigma_x\equiv \frac{\displaystyle\int_{\mathbb R} |x|\;|\psi(x)|^2\,\mathrm ...
15
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(...
- 2,913
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 ...
- 53.6k
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, ...
- 94.5k
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