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0

It boils down to a matter of convention. Nothing stops you from choosing the annihilation operator to be $a^\dagger$. Still, in quantum field theory, you decompose e.g. a scalar quantity in plane waves $$ \Phi(x) = \int \frac{d^4 k}{4 \pi} \left( a(k) e^{ik\cdot x} + a^*(k) e^{-ik\cdot x} \right)$$ where obviousely the second part is the c.c. of the first ...


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The factor of $1/2\pi$ is an artifact of the normalization convention being used for the momentum eigenstates. To begin to see how this is so, let us note that the choice of normalization of a Dirac-orthogonal continuous basis completely determines the form of the resolution of the identity. Writing an arbitrary state $|\psi\rangle$ in a given ...


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Whether the upward direction will be taken as positive & the downward direction as negative or vice versa simply depends upon you. It does not bother any physics. Generally, the downward direction is taken as negative. The negative sign indicates the opposite direction only of what you have assigned the positive direction. In Principles of Physics by ...


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It depends on what direction you assign to be positive in your coordinate system. To avoid confusion, just remember which direction acceleration is acting and which direction you assigned to be positive.


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Well, things don't fall up, so gravity would be negative. The longer version: For classical mechanics, the direction of a coordinate system is often arbitrary. For free fall problems I generally choose down to be positive (especially when I'm not dealing with an initial velocity, but it's a matter of preference). What you do need to know, no matter what ...


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If up is positive, and gravity points down, then $a$ (acceleration due to gravity) would be downwards, so it will have a negative magnitude.


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I think you have a problem with Kirchoff's laws.Now it states that the potential drop across a closed loop is 0. Your txt seems to have taken the direction of I as the direction of electron flow.Now suppose you move along the wire anticlockwise(The txt has moved clockwise.) Across the resistor ,potential drops but you travel along the original direction of ...


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Furthermore, if PE decreases as r gets larger, then why does a pen gain PE as you lift it further away from a table and why does the PE decrease/convert to KE when you drop the pen? These are two entirely different contexts. In the first context, the potential energy of the system of two objects is considered. In this context, the magnitude of the ...


1

First of all, you wrote the equation for $U$ wrong; it should be $r$ instead of $r^2$ in the denominator. However, that typo isn't what the problem is. The problem is that you've overlooked the word "magnitude" in the question. If a negative number is changed to be a different negative number that's closer to zero, then the magnitude of the number has ...


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we take positive charge as a point charge because positive charge has a high potential and electrons are always move from lower potential to the higher potential.so , we use positive charge as a point charge


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Every time you look up "the" spherical mirror formula, it comes with a set of "where's". These define what each symbol stands for, and the sigh convention to use to distinguish the location of objects and images and the difference between concave and convex radii. You can find different-looking spherical mirror formulas, with (naturally) different sets ...



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