110,367 reputation
7162321
bio website motls.blogspot.com
location Czech Republic
age 41
visits member for 4 years, 2 months
seen 4 mins ago

Hi, I am a string theorist and a publicist.


3h
comment What experiment would disprove string theory?
Yes, we can calculate almost every physical observable in string theory that is well-defined, including many of those that were out of reach of the previous approximate theories. Tens of thousands of papers are full of such calculations.
18h
awarded  Nice Answer
1d
comment Representation of operator in “$a$-representation”
If by $X$, you mean something else than the usual (Hermitian) $X$, then you haven't given enough information to describe what $X$ actually is. In that case, everything you wrote indicates that $X$ is effectively $a$, up to a simple $c$-number phase redefinition that clearly changes nothing. If the phase is supposed to be an operator, then you haven't specified what the operators $\phi,X$ are supposed to be. Clearly, for any operator $\phi$, one may define $\exp(-i\phi)a$ and call it $X$. But what's the point here? Those operators will have no usual properties we associate with "phases" etc.
1d
comment Representation of operator in “$a$-representation”
You're mixing operator identities with some equations for some special eigenvalues that must be defined by additional formulae. Your formula $U^\dagger a U=X$ isn't an operator identity. Quantum mechanically, it's trivial to see that it can't be correct, either. Use any proof I gave you above, or any other. For example, $U^\dagger a U$ is clearly not Hermitian because its Hermitian conjugate is $U^\dagger a^\dagger U$. But $X$ is Hermitian. So the two sides cannot be equal as operators. Also, the very definition of $U$ is supposed to have $N,U$ as operators, but $\phi$ is a $c$-number constant
1d
comment Representation of operator in “$a$-representation”
Moreover, $X=\exp(-i\phi)a$ is a very unnatural formula in QM to be called a "rotation" of operators. Operators don't commute so $\exp(-i\phi)a\neq a\exp(-i\phi)$. By a rotation of an operator, we usually mean things like $\exp(i\phi)X\exp(-i\phi)$, the conjugation. Classically, the exponentials (phases) would completely cancel.
1d
comment Why are the proper time and the proper length not defined in the same frame of reference?
The proper time of a process P (that the object O, e.g. a train, undergoes) and the proper length of object O are defined in the same frame, namely the rest frame of O (in your case the train).
1d
comment Representation of operator in “$a$-representation”
If you want to introduce "polar coordinates" in the phase space, then $r$ and $\phi$ would be non-commuting, just like $X$ and $P$ are noncommuting. But for $r$ and $\phi$, the commutator has to become infinite near $r=0$. So in any basis, some matrix entries of either $r$ or $\phi$ have to be infinite. There is no good definition of the $\phi$ operator in particular. One may define $r$ as $\sqrt{H}$, the square root of the Hamiltonian, but the angle just isn't well-behaved. Effectively, you are trying to determine $X$ and $P$ at the same moment if you want to define $\phi$ near $r=0$.
1d
comment Representation of operator in “$a$-representation”
As I have already told you about thrice with three different proofs, no "angle" that would make your equation valid exists. Classically, you may write $a=\exp(i\phi)X$ for a complex number and a real number. But quantum mechanically, $X$ and $a$ becomes operators, and the uncertainty principle-related problems near $a=0$ prevent one from making $\phi$ a well-behaved operator.
1d
comment Representation of operator in “$a$-representation”
You are missing that $a$ is $(x+ip)/\sqrt{2}$. Classically, it becomes the "whole complex number" parameterizing the phase space, so unlike $P$, it is in no way a rotated $X$.
1d
comment Representation of operator in “$a$-representation”
@user41208 - what is $\phi$? It is easy to see that there is no well-behaved operator $\phi$ for which your $X=\exp(-i\phi)a$ holds. For example, the right hand side annihilates the ground state, a normalizable vector, because $a|0\rangle=0$. On the other hand, the left hand side annihilates no normalizable nonzero vector. Only the $\delta(x)$ wave function is annihilated by $X$, and this vector isn't normalizable (it has infinite norm).
1d
comment Representation of operator in “$a$-representation”
An overcomplete set of vectors is one such that each vector in the set may be written as a linear combination of the remaining ones. So there is a "greater number" of vectors in the set than what is needed for a basis.
1d
answered Relativity asymmetry?
1d
comment Representation of operator in “$a$-representation”
$X$ is in no natural way a "rotation" of $a$: the former is Hermitian, the latter is not, and this property can't be changed by a rotation. Because $a$ is not Hermitian, the term $a$-representation is problematic. The set of $a$ eigenstates, the coherent states, isn't a basis: it is overcomplete. So there is no natural unique "matrix" representing an operator in this "representation".
Mar
25
awarded  Good Answer
Mar
24
awarded  Enlightened
Mar
24
awarded  Nice Answer
Mar
18
awarded  Good Answer
Mar
17
comment Do photons with a frequency of less than 1 Hz exist?
Sorry, the second half of the answer is silly. There is absolutely no problem in producing photons of less-than-one-hertz frequency. Just oscillate with a charge at a frequency less than one hertz and be sure that lots of electromagnetic waves are produced. The waves are composed of many photons - at such low frequencies, we actually produce a huge number of photons. What is hard is to "count" them or experimentally prove that the number is integer - it is not hard to produce them.
Mar
13
comment Why the CMB has not been dispersed so far?
Or alternatively, imagine that the box is as huge as the whole Universe and you're inside it - imagine somewhere near the center. The radiation gets diluted as the box grows but you're still inside it and the radiation is still coming from all directions.
Mar
11
answered Decomposition of group representation using tensor method