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## Hot answers tagged heisenberg-uncertainty-principle

23

If I understand you correctly you have an experimental setup like this. (image partially taken from The Physics of Springs) You measure two things: the $x$ position given by the position where the fluorescent spot appears on the screen $p_y$, the $y$-component of momentum given by the compression of the spring when hit by the particle You claim there ...

10

Yes, indeed, the average lifetime will tend to the normal distribution, as required by the Central limit theorem. A caveat: one has to distinguish between the distribution of lifetime and the distribution of the average lifetime. Distribution of lifetimes can be taken as exponential: $$w(t)=\frac{1}{\tau}e^{-\frac{t}{\tau}},$$ where $\tau$ is the mean ...

8

A particle's lifespan is has a memoryless and hence exponential distribution. If $\tau$ is the mean lifetime, the lifetime has CDF $1-e^{-t/\tau}$ (so the half-life is $\tau\ln2$) and PDF $\frac{1}{\tau}e^{-t/\tau}$ for $t\ge0$. Therefore, the proportion of surviving particles after a time $t$ is on average $e^{-t/\tau}$, and the mean lifetime of one ...

6

Thomas Fritsch's answer covers the fact that your experiment doesn't probe the essential problem. So what is the essential problem? Your measurement system can record whatever it records to whatever precision it's capable of. The HUP constrains your ability to predict what it will record. If you prepare many electrons in exactly the same way, so their ...

6

There are two issues with your question. First the uncertainty principle does NOT state that there is no state where $A$ has a definite value. It states (somewhat loosely) there is no state for which $A$ AND $B$ have simultaneously definite values. The uncertainty principle is a bound on the product of variances of hermitian operators, not on possible ...

5

When an electron hits the screen, it will give us a spot of diameter $\Delta x$, whereas the compression of the spring is also measured with a certain precision, which will limit the precision of measuring momentum to $\Delta p$. The Heisenberg uncertainty principle then says that $$\Delta x \Delta p \geq \frac{\hbar}{2}.$$ There is nothing special about ...

3

As a simpler case, think back to one-dimensional single-particle quantum mechanics. In that context we have an uncertainty relation $\Delta x \Delta p \geq \hbar / 2$. And in that case, although we like to write things like "let $| k\rangle \in \mathcal{H}$ be a momentum eigenstate" this is sort of a polite fiction: really the Hilbert space $\... 3 You are assuming that the screen is a classical object with a perfectly well-defined position, a perfectly well defined distribution of detection points, and that the spring has a perfectly well defined length. In reality, the screen and the spring are made of electrons and ions, all of which have imperfectly defined positions and momenta in accordance with ... 3 No. The uncertainty principle is a general relation between expectation values of quantum mechanical operators - the Robertson-Schrödinger uncertainty relation. The string length is not an operator in string theory, it's just a constant$\propto \sqrt{\alpha'}$, where$\alpha' = \frac1{2\pi T}$for the tension$T$that appears as the free multiplicative ... 3 Then, it is possible to prove the uncertainty principle with mathematical precision. It says that there is no state where$A$(or$B$) has a definite value This is not what the uncertainty principle says. However, if we replace your "or" in the parenthesis by "and", then we will get a statement that uncertainty principle does say. The ... 2 It is true that$\langle \alpha\vert L_z\vert\alpha\rangle=0$if$\vert\alpha\rangle$is an eigenstate of$L_x$OR an eigenstate of$L_y$. Indeed, if it’s an eigenstate of$L_x$then$\Delta L_x=0$so the uncertainty relation yields $$0\times \Delta L_y=0\ge \frac{1}{2}\vert\langle \alpha\vert L_z\vert\alpha\rangle\vert \tag{1}$$ which implies the right ... 2 According to the closure relationship we have for any two components of the angular momentum: $$[\hat{J}_i,\hat{J}_j]=i\hbar\epsilon_{ijk}L_k\tag{1}$$ where$\epsilon_{ijk}$is Levi-Civita symbol and the sum is understood over the repeated indexes. A non-zero commutator implies that there is no common basis of eigenvectors. That said, common eigenstates can ... 1 Unfortunately, the words such as uncertainty, resolution, accuracy, etc., are used in everyday language do not mean the same as in physics or in engineering, and because of that these concepts lead to confusion. Anyhow, there really are two completely different concepts frequently confused, namely accuracy and resolution. Accuracy is essentially the standard ... 1 It depends on how you define "Fock space". Photon sources actually create "wave packets". Basically you have $$\hat A^{\dagger}(\alpha)=\int_0^{\infty} d\omega \alpha(\omega)\hat a^\dagger(\omega)$$ where$\hat a^\dagger(\omega)$creates a photon of frequency$\omega$and$\alpha(\omega)\$ is a distribution of frequencies in your wave ...

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