# Tag Info

### 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 ...
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, ...
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 $$\mathbb E(A)=\sum_i a_i\,p(a_i)$$ ...
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 ...
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,...

### 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 ...

### 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 ...
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 ...
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 $\mathcal H$...
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 ...

### 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 ...

### 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 ...

### 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....
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 ...

### 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) ...

### 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 ...
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 ...

### 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 ...

### 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 ...

### 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/...

### 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 ...

### 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 ...

### 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 ...

### 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, ...
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 ...

### 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 ...
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 ...

### 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(...