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I know that the Bandwidth Theorem (BT) and the Heisenberg Uncertainty Principle (HUP) are basically the same thing, and stem from the fact that for operators $A,B$, we have:

$$\Delta A \Delta B \geq \frac{1}{2} | \langle \psi |[A,B]|\psi\rangle.$$

The HUP is obtained using the fact that $[x,p] = i\hbar$. In my class, we were asked to prove this inequality, but were given without proof the value of $[x,p]$.

I've now been asked to prove the bandwidth theorem, which states that

$$\Delta f \Delta t \geq \text{some positive constant}$$

I assume that $f$ is frequency and $t$ is time. It seems that this theorem is folklore, because I haven't been able to find any rigorous proofs. It seems intuitive, especially after watching this 3B1B video, however I have some questions about how to prove this:

  1. What even are $f$ and $t$? Functions? I know that the formula for $\Delta A^2$ is given by $\langle \psi | A^2 |\psi\rangle -\langle \psi | A |\psi\rangle^2 $. It's unclear how to compute this for $f$ or $t$, since I don't see how they can "operate" on $\psi$. I'm aware that we can use $$\langle f_1(x)|f_2(x)\rangle := \int_\mathbb{R} \bar{f_1}(x)f_2(x) dx$$ as an inner product.
  2. Similarly, how was $[x,p] = i\hbar$ obtained?

I'm sure that the answer has something to do with a Fourier series, but my knowledge of them is shaky (basically just know the definitions), and most discussions of this theorem that I've found have been handwavy to say the least.

For what it's worth, I would still like to figure this problem out for myself (I'm not asking for a full solution), but I need to understand more about what the BT is actually stating before I can try to prove it.

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Where $t$ and $f$ enter

You are correct: $t$ in quantum mechanics is not clearly an observable and $f$ is mostly intelligible only by the Einstein relation that $E = h f$ as the observable $\hat H/h,$ which is somewhat unsatisfying in trying to derive an uncertainty principle between them. This is fundamentally a Fourier-space argument about a time signal but wavefunctions are not time signals, they are distributions that may evolve over time.

As you might expect in terms of teachers who pay considerable attention to detail, Griffiths's QM textbook stresses this point considerably in section 3.5.3 where he derives this new point, saying,

My purpose now is to derive the energy-time uncertainty principle, and in the course of that derivation to persuade you that it is really an altogether different beast, whose superficial resemblance to the position-momentum uncertainty principle is actually quite misleading.

Indeed the general uncertainty principle that you have already come to contains the thrust of his derivation, which is that if the system is going to be observed to change, this change must be witnessed by some observable $\hat A$. It changes at a rate $\frac{d\langle A\rangle}{dt}$ and to be sure of the change you want it to have changed by an amount proportional to $\Delta A$, which motivates a definition of a characteristic time length over which the system has changed,$$\Delta t = \frac{\Delta A}{d\langle A\rangle/dt}.$$

Once you have this definition you can directly see the application to your case because by the chain rule, $$\frac{d\langle A \rangle}{dt} = \langle \partial_t \psi |\hat A|\psi\rangle + \langle \psi |\partial_t \hat A|\psi\rangle + \langle \psi |\hat A|\partial_t \psi\rangle.$$ Since $i\hbar |\partial_t \psi\rangle = \hat H |\psi\rangle$ and we generally would want $\partial_t \hat A = 0$ so that the thing we're measuring to witness the change is not itself changing with time, we are left with $$i\hbar~\frac{d\langle A \rangle}{dt} = \langle A H - H A\rangle.$$ I like to call this the "ah-ha" relation because I mostly just tutor individual students and occasionally they groan, but they remember it. It's terribly useful especially when one gets a little further and starts to do all of quantum mechanics in this way, the so-called "Heisenberg picture" of quantum mechanics. But anyway, this is just your commutator above, so one rather directly has $$\Delta E~\Delta A \ge \frac12 ~\hbar~ \left| \frac{d\langle A\rangle}{dt} \right|,$$ and with $\Delta f = \Delta E/\hbar$ one then has the desired relation $\Delta f ~\Delta t \ge 1/2.$ It says that wavefunctions with very precisely defined energies and therefore very precise frequencies cannot change in any observable way, except as the proof indicates above -- they can be "seen to change" by observables that themselves are not constant in time.

With that said there is also a general Fourier relation between $t$ and $f$ for signals just like there is between $x$ and $1/\lambda$ for spatial signals: as you squash a signal into a smaller and smaller time/space "domain", you necessarily make it harder to resolve anything but the highest frequencies as those can still have many repetitions across the domain. As a result the distribution in the conjugate Fourier space widens whenever you do anything to try and narrow the distribution in the base space, and vice versa -- if you want a signal that has a very well-defined wavelength then it needs to consist of many, many repetitions of that wavelength, the more the better; and so it needs to occupy a very large amount of your base space. You can even view the $x/p$ relation and the $t/E$ relation as special relativistic analogues of each other. But in nonrelativistic quantum mechanics this relation persists -- but it is not meaning quite the same thing in the same way.

The canonical commutation relation

In many contexts we would say that $[x, p]=i\hbar$ is just a fundamental axiom of the theory. What it means is that $p$ as an operator takes the function of $-i\hbar\partial_x$, since $[x, \partial_x] = -1.$ In turn this is a deep way of trying to induce the de Broglie relation that $p = h/\lambda.$

That is made more clear when one realizes that one is dealing with Fourier transforms of probability distributions over space. So one starts from some $\psi(x)$ and constructs its Fourier transform, $$\psi[K] = \int_{-\infty}^{\infty} e^{-2\pi i~K~x}~dx~\psi(x).$$Note that when we use this $2\pi i$ factor in the Fourier transform we're automatically normalized, but we do have to remember that $K = 1/\lambda$ (as opposed to how physicists normally think about $K$-space which is that $k=2\pi/\lambda$). Normalization just means that: $$\int_{-\infty}^\infty dK~ \psi^*[K]~\psi[K]=\iint_{\mathbb R^2}dx~dx'~\psi^*(x')~\psi(x)~\delta(x-x')=1,$$because $\int_{\mathbb R} dK~e^{-2\pi i ~K~u} = \delta(u),$ where $\delta$ is Dirac's delta-function.

But when we want to perform the multiply-by-$K$ operation in Fourier space that now has a direct interpretation in position space because we can just write it using the inverse Fourier transform as,$$\hat K \psi(x) = \int_{-\infty}^{\infty}~e^{2\pi i~K~x}~dK~K~\int_{-\infty}^{\infty} e^{-2\pi i~K~x'} ~dx'~\psi(x'),$$ and one can simply rewrite that internal $K$ as an operation acting upon the exponential, which is the only thing that depends on $x$, $$\hat K \psi(x) = \int_{-\infty}^{\infty}~dK~\frac{1}{2\pi i}\partial_x ~e^{2\pi i~K~x}~dK~\int_{-\infty}^{\infty} e^{-2\pi i~K~x'} ~dx'~\psi(x').$$ Commuting this operation on $x$ out of the integral on $K$ and using $\hat p_x = h~\hat K$ as our de Broglie relation gives the formula $\hat p_x = -i\hbar \partial_x$ more directly. So if one looks in Fourier space it is obvious that $\hat p = h~\hat K$ as suggested by de Broglie must map to the equivalent $\hat p = -i\hbar \nabla.$

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its pretty simple its energy time unicertanity. basically assume that we have $$H|n\rangle=E_n|n\rangle$$ where $|n\rangle$ is an eigen state of the hamiltonian with energy $E_n$.

so the next question is this, if the an electron is in state $|n\rangle$ at the begning in which state it will at time $t$. The answer is

$$|n,t\rangle=e^{iE_nt/\hbar}|n\rangle$$

now we know that $$E_n=h f_n=2\pi\hbar f_n$$ thus we have $$|n,t\rangle=e^{i2\pi f_nt}|n\rangle$$

now lets go to the general state $$|\psi\rangle=\sum_n C_n|n\rangle$$ and

$$|\psi,t\rangle=\sum_n C_ne^{i2\pi f_nt}|n\rangle$$

so now comes the bandwith part

the timedependent exponential parts are just sinusoidal waves, so by using this a you can find a relation ship between the bandwith of $E$ that is the form $C_n$ with band with of $t$ that is the band with of the wave packet $$A(t)=\langle\psi|psi,t\rangle=\sum_n |C_n|^2 e^{i2\pi f_nt}$$

example

iff $C_0=1$ then $ |A(t)|^2=1$ so the state will stay at the same eigen state for ever it is stationary band with of $A$ is infinite while bandwith of $C_n$ in other words $f_n$ is zero if you increase the former latter will decrease. if C_n have a large band with then the state will rapidly move away from the initial state as time goes on.

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