# Tag Info

3

Your everyday physics intuition is leading you astray. The kinetic energy of the incoming particles can't just disappear, so they can't just come to a stop unless they can transfer the kinetic energy into some other degree of freedom. If you collide two balls of clay they will come to a stop, but only because the kinetic energy can go into deforming and ...

1

The relevant physics argument here is at heart a scaling argument, which is why it would be more instructive to do this exercise in natural units where $\hbar = 1$. If we consider objects of uniform density then to study what happens if these objects become extremely large, we can define a new scale to measure their positions and momenta. We define new ...

2

The $\Delta$ values in the uncertainty relation tell you the errors (or uncertainties) in the position and momentum, not the actual/measured values of these. For a "normal" sized body, with momentum 1kgm/s, there may be an error of say $\Delta p =\pm0.01$kgm/s for example. Follow through with this value in the error for position and see what you get...

0

I can only answer your first question due to time constraints: To me the phrase "interaction cannot be allowed for" is a roundabout way of saying they cannot determine the photon's exact momentum for use in the related equations.

2

The uncertainty principle is still true in its usual form, but it refers to your knowledge of the state. Suppose your state is just one electron, you can confirm this if you observe the system for a time $\Delta t$ and you don't see additional particles. However, due to the uncertainty principle you can only measure particles that increase the energy of the ...

9

There are many steps: Step 1, select a state $\Psi$. Step 2, prepare many systems in same state $\Psi$ Step 3, select two operators A and B Step 4a, for some of the systems prepared in state $\Psi$, measure A Step 4b, for some of the systems prepared in state $\Psi$, measure B Now if you analyze the results, assuming strong (not weak) measurements then ...

4

One of the main problem about the uncertainty principle as it is usually told in quantum mechanics is that it is always told in the historical context, giving what Heisenberg thought about it or Feynman. For once (at least), this is not very clever. In today's literature, we distinguish different types of "uncertainty relations", based on what they actually ...

6

You cannot get both $\langle \psi \rvert A \lvert \psi \rangle$ and $\langle \psi \rvert B \lvert \psi \rangle$ if you only have one state $\lvert \psi \rangle$, or if you always measure $A$ on your states. What you need to do to check the uncertainty principle experimentally is preparing an ensemble of identical states and then measure $A$ on one half and ...

1

In quantum mechanics, a system is represented by a wave function. The wavefunction is a sort of blob spread out over $x,p_x$ space and $\delta x, \delta p_x$ represent the size of the blob. At a given time the wavefunction of a system has to be spread out in x if it is highly peaked in $p_x$ and vice versa because $\delta x \delta p_x \geq \hbar$. The ...

-1

Yes, it would be a violation of the uncertainty principle. Therefore is represents an argument against faster-than-light communication which allows sending info to the past and receiving info from future. For example: in a bipartite experiment with suitably entangled pairs of particles, if Alice measures on her particle $p_y$, and $x$ and Bob measures ...

1

The $\sigma_A$ on the LHS is defined as $$\sigma_A = \omega(A^2 - \omega(A)^2I),$$ where $\omega$ is a state (so $\sigma_A$ also depends on $\omega$!) and the way this is defined by von Neumann is by taking a statistically relevant ensemble of very same copies of the very same system in the very same state $\omega$ and repeat the measurement of $A$ and $A^2$ ...

0

Maybe statistics and the HUP are conjoined in an inseparable way in QM, * Yes, they are, because a quantum system has the property that some observables are non-defined. The standard QM says that this property is not a flaw in our measurements, but an intrinsic property of the quantum system itself. Let $x$ and $p_x$ be the two non-commuting ...

2

The uncertainty product is bounded from below by the expectation value of the commutator of the relevant observables. If $A$ and $B$ are any two observables, then the generalized Heisenberg uncertainty relation reads as $$\sigma_A\sigma_B \geq \frac{1}{2}\vert \langle[A,B]\rangle\vert .$$ For the case of position - linear momentum pair, the commutator is ...

3

The Heisenberg uncertainty principle in the most general form $$\Delta_\omega(A)\Delta_\omega(B)\geq\frac12|\omega([A,B])|$$ depends on the state $\omega$ on which it is evaluated. In the special case of the canonical commutation relations $[q,p]= i\hbar I$, $\omega(I)=1$ for any state and therefore the RHS reduces to a constant. For more general commutators ...

9

There is a 1961 paper by Aharonov and Bohm on this subject, in which there is defined, among other things, a characteristic time for an operator's expectation value to deviate significantly, measured by the initial dispersion in that operator. This result is essentially a theorem we will prove here (Theorem 2). Let $\mathcal{S}$ be a Hilbert space, $A$ a ...

0

Time is not an operator, s.t. this type of uncertainty relation differs from the usual form, in which participate operators. (There are though attempts to define time as an operator in time-arrival problems). And though, in nuclear physics we are well acquainted with unstable (radioactive) nuclei. The unstable nuclei are so-called resonant states, and the ...

3

Ok, so the Energy-Time "Uncertainty Principle" is often way misrepresented, but it does mean something. In $\Delta t \Delta E \geq \frac{\hbar}{2}$ the $\Delta t$ represents the amount of time is takes for the energy expectation value to change by one standard deviation. It does not represent a measurement. Note: This is explained in Griffith's "Quantum ...

1

I believe that the relation between admissible values of an observable and Heisenberg uncertainty principle is best seen in the axiomatic formulation of quantum mechanics. Starting from the postulate that every quantum mechanical system is described by a C*-algebra $A$, an admissible value $\lambda$ for the observable $O\in A$ is the value of a state ...

1

Heisenberg uncertainty is, in its general form as Robertson-Schrödinger uncertainty, given by $$\sigma_A \sigma_B \geq \frac{1}{2} \langle [A,B] \rangle$$ for any two observables $A,B$. Now, in an eigenstate $\lvert a \rangle$ of $A$, $\sigma_A$ vanishes, but also $\langle [A,B]\rangle = \langle a \rvert(AB - BA) \lvert a \rangle = \langle a \rvert( a B ... 0 Heisenberg Uncertainty Principle is always applicable. The energy eigenstate will remain so for an infinite time as long as a perturbation leads it to decay. 2 I would say that an answer is that length of wave packet and width of spectrum are related by:$\Delta\omega \Delta t\approx 1\$ "Width of spectrum" here is characteristic range of frequencies that signal contains, that is width of Fourier transform of the signal. Infinite sine wave contains only 1 frequency, that is its spectrum/Fourier transform is ...

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