Hi everybody,
I encountered this question as an example of what a google applicant is asked:
You are shrunk to the height of a 2p coin and thrown into a blender. Your mass is reduced so that your density is the same as usual. The blades start moving in 60 seconds. What do you do?
It is...
Hi togehter.
I encountered the following problem:
The timeordering for fermionic fields (here Dirac field) is defined to be (Peskin; Maggiore, ...):
T \Psi(x)\bar{\Psi}(y)= \Psi(x)\bar{\Psi}(y) \ldots x^0>y^0
= -\bar{\Psi}(y)\Psi(x) \ldots y^0>x^0
where \Psi(x) is a Dirac...
Thanks again. But that's what i meant.
It is unlikely but not impossible. And a very poor lower bound doesn't justify to call \Delta p the typical momentum.
It seems as if one could conclude the mean value (which i assume is the typical value) from the standard deviation. But there's no...
Thanks, but i think it's meaningless to give a lower bound and speak of typical momenta.
The uncertainty can be a very small fraction of the actual momentum or vice versa, so i can't believe they meant it this way.
They say:
"The uncertainty principle gives an estimate for the typical electron momenta when they are confined to such a linear domain." Then they give the above formula.
But all they get from the uncertainty principle is the uncertainty \Delta p which is not the typical momentum. In my...
Sorry for bumping, but i can't belive that nobody who did some lectures on particle physics ever encountered this (or a similar misuse of the uncertainty principle) problem. I would be very glad for an answer.
Greetings.
They conclude that the momentum of the electron is \approx \hbar/\Delta x , so it surely isn't zero. They want to calculate the typical momentum of an electron in an atom, it would not be very helpful to chooce such a reference frame, because the answer would be that the typical momentum is...
Hi together ...
In many textbooks on particle physics i encounter - at least in my mind - a misuse of the Heisenberg uncertainty principle.
For completeness we talk about
\Delta p \Delta x \geq \hbar/2
For example they state that the size of an atom is of the order of a few Angstroms...
Hi.
The whole thing is rather sloppy and mathematically not very well justified by Dirac.
If you have an operator A in the Hilbert space the eigenvalues and eigenkets of the operator satisfy
A |a\rangle = a|a\rangle
where a is an eigenevalue if you have a continuum of eigenvalues in an (real)...
O.k. so everything in the exponent multiplied by t except the -i is the frequency \omega. As E=\hbar \omega the negative energy comes from the negative frequency. Mhh i think i probably got it although there are many cases i remember where the time dependence was \exp(i \omega t) and no one...
my point is:
you say that there are negative energy solutions because, as you showed, the general solution bears a plus AND a minus sign in the exponent. but it seems to me it would be the same to say the energy can be imaginary because there is an i in the exponent. i simply don't get the...
thank you. i also heard this statement, but it's the same story i think. only this time the sign is carried over to the paramter t instead of the energy.
i'll try to make my point clear:
although the solutions are
\Psi(t,\vec{x})=\exp(\pm i (- E t + \vec{p} \cdot \vec{x}))
you don't...
yes, but as i mentioned earlier, why does this sign carry over to the energy? why don't say the energy is positive but the functional dependence, to get linearly independent solutions, is exp(- ...) respectively exp(+ ...)?
thanks for your help and please excuse my problems understanding that...
Thanks. Thats a linear superposition of the fundamental solutions i cited above, but as you wrote \omega and thefore the energy is positive and there is no need for negative energy states called antiparticles. Or did i misunderstand something?
edit:
Back when they dealt with the Klein-Gordon...
that was my point ...
I know that they have been found experimentally. But it is always quoted as a great triumph of theoretical physics that antiparticles were predicted on grounds of the Klein-Gordon equation or the Dirac-Equation respectively.
Hi together ...
I wonder how one can deduce the existence of antiparticles from the Klein-Gordon equation.
Starting from (\frac{\partial^2}{\partial t^2} - \nabla^2 + m^2) \Psi(t,\vec{x})=0
one gets solutions \Psi(t,\vec{x})=\exp(\pm i (- E t + \vec{p} \cdot \vec{x})) leading to E^2=p^2 +...
Thanks, but in the statement i quoted the domain of A(t) isn't U(t)D(A) but U(t)^+ D(A) and then we have (U(t)^+ \Psi, U(t)AU(t)^+ U(t)^+ \Phi). Is this a typo?
Hi together ...
I encountered the following statement:
Operator A is self adjoint on D(A) then A(t) \equiv \exp(iHt) A \exp(-iHt) is self adjoint on D(A(t)) \equiv \exp(-iHt) D(A).
H is self adjoint, so that exp(...) is a unitary transformation. But why does the domain transform this way...
hi.
i'm reading "quantum mechanics in hilbert space" and a don't get a basic point for bounded operators.
def. 1 a set S in a normed space N is bounded if there is a constant C such that \left\| f \right\| \leq C ~~~~~ \forall f \in S
def. 2 a transformation is called bounded if it maps...
Hi.
I just calculated the quantum mechanical harmonic oscillator with a driving dipole force V(x,t) = - x S \sin(\omega t + \phi)
I used the Floquet-Formalism. Then I calculated the mean expectation value, in a Floquet-State, of the Hamiltonian over a full Period T indirectly by using the...
Hi together...
When reading Sakurai's Modern Quantum Mechanics i found two problems in the chapter "Approximation Methods" in section "Time-Independent Perturbation Theory: Nondegenerate Case"
First:
The unperturbed Schrödinger equation reads
H_0 | n^{(0)}\rangle=E_n^{(0)}...
Great.
Thanks for your answers. i hope i got it (at least partially).
@Fredrik:
Thanks for your rigorous explanation. But the Hamiltonian i wrote above can't be brought in this "non-interacting" from.
Hi.
I found in a book on quantum optics (Vogel, Welsch - Quantum Optics) and also in a lecture script the following statements.
A System is composed of two subsystems (say A and B) which interact. So the total Hamiltionian is H_{AB}= H_A + H_B + H_{int}.
Nontheless the states of H_{AB}...
Thanks for your replies.
I found the problem. The identity only holds for the special case of creation-/annihilation-operators, where the (anti-)commutator for fermions or bosons resplectively is zero.
thanks and greetings.
Hi all.
I found the following identity in a textbook on second quantization:
([a_1^{\dagger},a_2^{\dagger}]_{\mp})^{\dagger}=[a_1,a_2]_{\mp}
but why?
([a_1^{\dagger},a_2^{\dagger}]_{\mp})^{\dagger}=(a_1^{\dagger}a_2^{\dagger}\mp a_2^{\dagger}a_1^{\dagger})^{\dagger}=a_2a_1\mp a_1a_2...
I thought i had a basic to intermediate understanding of quantum physics and group theory, but when reading hamermesh's "group theory and it's application to physical problems" there's something in the introduction i don't understand.
first of all, i know the parity (or space inversion)...