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2

If all you want to know is the voltage at one instant, you could indeed simply add the x-components. If you want to know the amplitude of the resulting sinusoid, you need to do complete vector addition (both x and y coordinates) to get the full amplitude (43.6 V. in your example). This will tell you not just what the voltage is at one instant in time, but ...


0

Let's say $\Phi$ is a delta function, $\Phi(k)=\delta(k-k_0)$. Presumably, you want this to be an eigenstate of the momentum operator with momentum $\hbar k_0$. With the convention you've chosen, we can convert this to a real-space wavefunction (I'm ignoring normalization for convenience): $$ \Psi(r)= \int dk \delta(k-k_0)e^{ikr}=e^{ik_0 r} $$ We can then ...


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In dV=-E.dr, E is electrostatic electric field and the negative sign implies that the electric potential V decreases with increasing electric field E. This is the significance of this negative sign. In E is not electrostatic, but induced electric field. This field is induced due to time-varying magnetic fields or motional electromotive force causing a ...


1

The three generators of right-handed spinor rotations are given by $\left\{- i\sigma_x,-i\sigma_y,-i\sigma_z\right\}$, see for instance Peskin & Schroeder page 44, and the rotation matrix for a spinor rotation over an angle $\phi$ around a unit vector $\hat{s}$ is given by: $R~=~ \exp\left(-i\frac{\phi}{2}~\hat{s}\cdot\vec{\sigma}\right) ~=~ ...


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Maybe this is a bit too general an answer but I hope this helps someone in the future. Here is the lenses makers equation write explicitly with the sign conversion terms, $$ \frac{1}{f} = \left(n-1\right)\left( \frac{d\left(n-1\right)}{\beta_1\beta_2n\bigl|R_1\bigr|\bigl|R_2\bigr|} + \frac{\beta_1}{\bigl|R_1\bigr|} - \frac{\beta_2}{\bigl|R_2\bigr|} \right) ...


0

There are different sign convention in optics. Once you accept one sign convention you should make sure that the answer is in the same convention. Application of the signs twice guarantee the same. How this magic happening? At the time of derivation you applaying the signs for U,V,F thar guarantee that U,V,F are in our convention. At the time of solving ...


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No, the easiest example is the spin of a fermion (electron, muon, etc...) which is always half integer.


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Quantum mechanics tends to quantize things, forbidding certain "continuous" solutions that would otherwise exist in favor of "discrete" solutions. Whenever those discrete solutions exist, you can assign some integer "quantum numbers" to a system and view solutions as an overlay of those numbers. A simple answer where they don't exist is if you have the ...


5

No, they are not. As ACuriousMind pointed out above, quantum numbers are just convention. They may be integers, real numbers, whatever depending on the problem discussed. Note that for example in quantum optics, coherent states $|α\rangle$ are defined in terms of complex quantum numbers, $α\in\mathbb C$. How they translate to actual physical quantities ...


0

There really is not a unique way to define the ladder operators, especially with the constants in front. If you change them from what one book has, you might match another book. Those differences will show up in different normalization constants for "next" states that the operators create. Eventually, you will use your operators to find the algebraic form ...


0

You are correct, but one place where it tends to bite you is in Fourier transforms. With Fourier transforms you really want the inverse transform to have $+i$ in it, so that $\delta(k - k_0)$ becomes $e^{i\,k_0\,x}$ when you transform back to position space. (It doesn't have to be, but that's the easiest to remember: the delta at $k_0$ produces ...


2

There is no physical significance because there are two equally good square roots of negative one, i.e. $\pm i$ and unlike $\pm\sqrt 2$ there is no way to tell the roots apart. So what you call $+i$ and what I call $-i$ could actually be the same. What matters is relative signs, for instance $e^{i(kx-\omega t)} is a wave traveling to the right. You ...


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The important thing is the relative sign between the potential and the Laplacian. Otherwise there are two square roots of $-1$ namely $\pm i.$ You could write $j=-i$ and then you have two perfectly good square roots of $-1$ you can use $i$ or you can use $j$. For square roots of positive numbers you get things like $\pm\sqrt 2$ and there is a way to pick ...



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