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    Shrunk and thrown into a blender

    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...
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    Product of Dirac Spinors

    Thanks a lot ...
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    Product of Dirac Spinors

    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...
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    Uncertainty principle

    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...
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    Uncertainty principle

    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.
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    Uncertainty principle

    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...
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    Uncertainty principle

    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.
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    Uncertainty principle

    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...
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    Uncertainty principle

    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...
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    Representing a vector in terms of eigenkets in a continuos basis

    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)...
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    How one can deduce the existence of antiparticles

    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...
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    How one can deduce the existence of antiparticles

    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...
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    How one can deduce the existence of antiparticles

    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...
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    How one can deduce the existence of antiparticles

    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...
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    How one can deduce the existence of antiparticles

    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...
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    How one can deduce the existence of antiparticles

    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.
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    How one can deduce the existence of antiparticles

    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 +...
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    Self adjointness and domain

    Thanks a lot for the hint. So it was an error in the book. For the record: B. Thaller - The Dirac Equation. Section 1.2.2. Greetings. Tommy
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    Self adjointness and domain

    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?
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    Self adjointness and domain

    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...
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    Bounded Operators

    Great! Many thanks to both of you.
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    Bounded Operators

    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...
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    Driven quantum mechanical harmonic oscillator

    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...
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    Stationary Perturbation Theory

    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)}...
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    Direct Product Basis for Interacting Systems

    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.
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    Direct Product Basis for Interacting Systems

    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}...
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    Adjoint of commutator

    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.
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    Adjoint of commutator

    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...
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    Invariance of Schrödinger's equation

    hi all. thanks for your quick answer. i'm going to consult landau and lifgarbagez.
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    Invariance of Schrödinger's equation

    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)...
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