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3

In the Heisenberg picture, one simply has $$A(t) = \exp(-Ht/i\hbar) A(0) \exp(+Ht/i\hbar)$$ The Hamiltonian $$H = \frac{E_1+E_2}{2}\cdot {\bf 1} + \frac{E_1-E_2}{2} \cdot \sigma_z$$ while $$A(0) = a\sigma_x$$ The term in $H$ proportinal to ${\bf 1}$ cancels in $A(t)$ so we have  A(t) = a\cdot \exp(-(E_1-E_2)t\sigma_z/2i\hbar) \sigma_x \exp(+(E_1-E_2)...

22

You don't. You actually hear the high frequency notes from headphones. The bass really doesn't travel at all well, but the attack noise from the drum or bass guitar is what leaks from headphones. This is why on the tube you hear "tsss tsss tsss tsss" and very little else. From @leftaroundabout's answer on the post that valerio92 linked: Normal ...

0

"does the shift directly depend on being above or below resonance?" Phase of reflected light will depend on which side of resonance laser is(check out cavity phase response). "I would have expected you to mix the output signal with the initial laser signal, such that the shift in phase could be determined and hence accounted for." You could in theory do ...

2

Let's back up. How do you know that a monochromatic wave of frequency $\omega$ doesn't contain any component at $\sqrt{2} \omega$? It's because these two frequencies are not commensurate: if you plot $\sin(\omega t)$ and $\sin(\sqrt{2} \omega t)$ they'll have no clear relation. The peaks of one look like random points in the other. Then it's clear their ...

2

To say that a wave, say with amplitude given by f(t), has a period $T$ means that not only $f(t+T) = f(t)$, but also that $T$ is the smallest value that has this property. Given that $f(t+T) = f(t)$ then it follows that $f(t + nT) = f(T)$ where $n$ is an integer (for example $f(t+2T) = f(t+T+T) = f(t+T) = f(t)$).

2

A sine wave at $\omega$ is periodic in $\omega /2$, but that does not mean that a sine at $\omega$ has frequency components at $\omega /2$. Another way of looking at it is noting that you cannot synthesize a sine at $\omega$ as the sum of sines of other frequencies.

1

To keep things mathematically both precise and simple, let's stick with the discrete Fourier transform. Signals are vectors of $N$ complex points, where $N$ is the dataset's length. The dimension of this vector space is $N$. In this setting, the Fourier transform is simply a resolution of a signal, thought of as a vector, into components with respect to a ...

1

I don't know what is happening there. It always worked for me. If you have large number of cycles with smooth variation i.e. large time scale with small time interval you will definitely see the frequency components. First of all matlab stores its frequency components like. 0 to $\omega_ {max}$ then $-\omega_{max}$ to 0. Hence if you want your zero ...

1

Based on some experience in music, I realized that the relation of the frequency of two pitches with equal steps say from C to D (whole step) or D to E (same whole step) is $f_2 = af_1$, where $f_2$ is the higher pitch, and $f_1$ is the lower, and $a$ depends only on the distance of steps from the lower to higher pitch. This means that the frequency ...

0

The context of the question is acoustics, but that is irrelevant. Essentially the same question could be posed in chemistry, biology, economics, etc. No physical laws or concepts can solve this problem. It is really a question about mathematics and the definition of "percentage difference." The same issue crops up regularly in mathematics homework ...

2

Not surprisingly, physicists have looked for variations in the speed of light as a function of frequency in vacuum. The state of the art in 1972 can be found in Z. Bay and J. A. White, 'Frequency Dependence of the Speed of Light in Space', Phys. Rev. D 5(4) 796-799 (1972). Using data from pulsar emissions (radio, visible, x-ray) and other sources (see paper),...

0

I like the following explanation. Although mostly mathematical, it illustrates a point Feynman tried to convey in his lectures. Energy is always conserved, and whenever it seems like it isn't you just need to look harder. Let's recall that in the context of classical particle mechanics one defines the energy of the system of $n$ particles as the sum of the ...

3

In a vacuum all frequencies and amplitudes of light travel at the same speed of c = 299 792 458 m/s. Frequency is equivalent to colour. Amplitude relates to intensity. When light travels in material mediums (air, water, glass, etc) it travels at a slower speed v < c which depends on frequency. The ratio of c/v is what we measure as the refractive ...

2

I believe that the frequency for the rotating disc of radius $r$ can be anything below $1/r$ (or $c/r$ in SI units). I think the picture you consider is flawed because you do not take into account the measurement of the speed by the observer on the disc. He would have to time the passing by 2 measuring roads at the previously known laboratory distance ($v = ... 1 Assuming that with "oscillation of the wave function" you mean that the state evolves in time with a relative phase$E/\hbar$as$\alpha\lvert 1 \rangle +\mathrm{e}^{\mathrm{i}E/\hbar t}\lvert 2 \rangle$, that's really an "accident" because given an energy, how else are you going to get a frequency in quantum mechanics out of it if not by dividing by$\hbar\$?...

0

The resonant frequency is equal to the natural frequency when no damping and no external force at all is applied to the system. When damping is applied so that now the decay time (decay of amplitude) is in effect, the resonant frequency decreases a little below depending on magnitude of damping.

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