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| bio | website | en.wikipedia.org/wiki/… |
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| location | ||
| age | ||
| visits | member for | 11 months |
| seen | 11 hours ago | |
| stats | profile views | 168 |
A Sufi prayer of love
O love, O pure deep love, be here, be now, be all.
Dissolve me into your stainless endless radiance,
Make me your servant, your breath, your core.
:- by sufi mystic Rumi
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Apr 9 |
comment |
Does the mass point move? Nice question +1. However it should be noted that (in framework of classical mechanics) only specification of the initial conditions is not enough to find out time evolution of an entity. One should also specify the force. |
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Apr 1 |
comment |
exponential potential quantization Did you mean $e^{bx}$ ? |
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Mar 22 |
revised |
A simple model that exhibits emergent symmetry? Included latex |
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Mar 22 |
suggested | suggested edit on A simple model that exhibits emergent symmetry? |
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Mar 14 |
awarded | Organizer |
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Mar 14 |
revised |
Did Hilbert publish general relativity field equation before Einstein? edited tags |
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Mar 13 |
comment |
Renormalization: Why is only a finite number of counter-terms allowed? As Josh answered a theory is called renormalizable if it requires only a finite number of counter terms for canceling infinities. Initially it was thought that non-renormalizable theories are in some sense bad in that they have no predictive power. However nowadays such theories too are seen with respect and are studied as "effective field theories". But still most physicists believe that a fundamental theory of nature should have the property of renormalizability. Here by fundamental I mean a theory written in terms of fundamental entities i.e those which can not be broken into smaller ones. |
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Mar 10 |
comment |
commutation of operator product expansion @Barefeg I made some changes. Actually you can conclude from two equations that O_2 and O_1 are equal. |
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Mar 10 |
revised |
commutation of operator product expansion added 44 characters in body |
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Mar 9 |
answered | commutation of operator product expansion |
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Mar 5 |
comment |
A universe of angular momentum? Relevant: is-all-angular-momentum-quantized? |
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Mar 5 |
comment |
A universe of angular momentum? @JohnRennie Sorry I didn't know that. Do you mean angular momentum observables can have states with continuous eigenvalues? |
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Mar 5 |
comment |
A universe of angular momentum? @Cheeku Observables in QM are matrices rather than functions so they do not commute in general(ie AB=!BA). In more simple terms suppose S is a physical system (say a particle) and A and B be two (quantum) observables associated with S. Say A= position and B=momentum. Now to know the value of A and B you have two choices - i) first measure A and then measure B ii) first measure B and then measure A. Classically both the ways are equivalent and will give the same result while for quantum systems these two ways of measurement are in general not equivalent and give different values of A and B. |
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Mar 5 |
comment |
A universe of angular momentum? However real essence of QM is not in the fact that angular momentum takes discrete values but in the fact that observables generally don't commute (whereas in classical mechanics they always commute). |
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Mar 5 |
comment |
A universe of angular momentum? Angular momentum is the most basic observable where QM differs from classical mechanics. In QM angular momentum takes only discrete values whereas (unless the particle is trapped in a finite region of space) momentum and position take continuous values in QM just as in classical mechanics. |
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Feb 21 |
comment |
In what sense is a scalar field observable in QFT? @user1504 I always had the viewpoint that observables have to do with symmetry of the theory. They form a representation of symmetry group on space of states. We can work in terms of fields or creation - annihilation operators or may be something else. The physical content of the theory is not in fields but in the space of states and the representation of the symmetry group on it. |
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Feb 21 |
comment |
In what sense is a scalar field observable in QFT? @user1504 I think from QM point of view its not electric and magnetic fields which are physical but the states in Hilbert space of U(1) YM theory. Quantum mechanically there can be no such thing as a "continuous (observable) field", we can only have states with discrete particle content (or their linear combinations). |
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Feb 21 |
comment |
In what sense is a scalar field observable in QFT? This i think is the correct answer +1. Why downvotes ? |
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Jan 30 |
comment |
Can ice have a higher entropy than water? "in the south pole, ice means higher entropy,.." Its true in following sense:- At south pole temperature of surroundings is below zero degrees i.e. <273K. Now consider ice and water in equilibrium at 273K. If water can lose some of its heat to the surrounding to form ice, then the entropy of total system ice+water+surrounding will increase. So the exact statement should be : "in south pole formation of ice is entropically favored". |
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Jan 30 |
accepted | Quantum mechanics on Cantor set? |
