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Yes. Thank you.

I am guessing time-energy uncertainty relation is the way to solve this. I solved the Schrodinger equation for both the regions and used to continuity at ##x=-a, 0,a## and got ##\psi(-a<x<0) = A\sin(\kappa(x+a))## and ##\psi(0<x<a) = -A\sin(\kappa(x-a))## where ##\kappa^2 = 2mE/\hbar^2##...

I do not understand how to use this fact actually. I am trying to eyeball some eigenfunction for ##e^{i\frac{d}{dx}}## but no success as of yet.

Okay. So I should simply state that the eigenfunctions are all those functions ##f(\phi)## which satisfy ##\operatorname{Ci}\left(\sin^{-1}\lambda f\right)/\lambda +C =\phi## and the corresponding eigenvalue is ##\lambda##. Is that right?

Thanks for your response. This is what I did $$\frac{df}{d\phi} = \sin^{-1}(\lambda f)$$ On integrating the ##df## integral using online integral calculator $$ \frac{\operatorname{Ci}\left(\arcsin\left(\lambda f\right)\right)}{\lambda} = \phi+C$$ It seems this cannot be expressed in terms of...

Here is what I tried. Suppose ##f(\phi)## and ##\lambda## is the eigenfunction and eigenvalue of the given operator. That is, $$\sin\frac{d f}{d\phi} = \lambda f$$ Differentiating once, $$f'' \cos f' = \lambda f' = f'' \sqrt{1-\sin^2f'}$$ $$f''\sqrt{1-\lambda^2 f^2} = \lambda f'$$ I have no...

Yes. Thanks a lot

Okay. Thanks a lot. One last question: We used an analog pressure gauge. So is the error in ##\Delta p=## least count of pressure gauge? Or should I take no error in ##\Delta p##? For my post #3 I assumed no error

The readings were taken by reducing the pressure by regular intervals of 10 torr and counting the number of fringes that moved outwards within that interval. I have already plotted the graph of ##m## vs. ##\Delta p## and can reasonably say my error in counting ##m## should be ##\pm 1##. I...

This is for the lab report I have to submit. ##n## is the refractive index. ##L## is the length of a gas chamber and ##m## is the number of fringes passed as the pressure in the gas chamber changes by ##\Delta p##. We are already given the error in ##L##. I performed the experiment and obtained...

I understand your point. But I am trying to teach myself GR and this exercise is in the first chapter of Carroll. He hasn't introduced curvature and stuff yet. Actually, I got the answer. It is ##\Delta \tau_B = \displaystyle\frac{L}{\gamma v}##. The mistake is in writing the coordinates and is...

Let us denote the events in spacetime before the trip has started by subscript 1 and those after the trip is over by subscript 2. So before the trip has begun, the coordinates in spacetime for A and B are ##A = (t_{A_1},x,y,z)## and ##B = (t_{B_1},x,y,z) = (t_{A_1},x,y,z)##. After the trip is...

Oh right! Thanks a lot!

Property (a) simply states that a second rank tensor that vanishes in one frame vanishes in all frames related by rotations. I am supposed to prove: ##T_{i_1 i_2} - T_{i_2 i_1} = 0 \implies T_{i_1 i_2}' - T_{i_2 i_1}' = 0## Here's my solution. Consider, $$T_{i_1 i_2}' - T_{i_2 i_1}' = r_{i_1...

I see it now. I was assuming the amplitude itself has dimensions of position. Thanks for helping @stevendaryl !

Yes! I had thought of that! But then am I not presuming that ##q## has dimensions of position?

The given lagrangian doesn't seem to correspond to any of the basic systems (like simple/ coupled harmonic oscillators, etc). So I calculated the momentum ##p## which is the partial derivative of ##L## with respect to generalized velocity ##\dot{q}##. Doing so I obtain $$p =...