$ \sigma _{H}\sigma _{Q}\geqslant \frac{h}{4\pi }\frac{d\left \langle Q \right \rangle}{dt}$  $\Delta E = \sigma _{H}$  $\Delta t = \frac{\sigma _{Q}}{d\left \langle Q \right \rangle / dt}$  $\Delta E \Delta t \geq \frac{h}{4\pi }$  Q is any observable  I know that $\Delta E$ represents the standard deviation of energy distribution, but what does $\Delta t$ represent precisely?  I read an answer saying "It is the average time of the expectation value to change by one standard deviation, but I don't understand this sentence, I need some clarification.
43012
In non-relativistic QM, the $\Delta E$ in the time-energy uncertainty principle is the limiting standard deviation of the set of energy measurements of $n$ identically prepared systems as $n$ goes to infinity. What does the $\Delta t$ mean, since $t$ is not even an observable?
71819
This is a more general treatment of the issue posed by  . After deriving the asymptotic distribution of the sample variance, we can apply the Delta method to arrive at the corresponding distribution for the standard deviation.    Let a sample of size $n$ of i.i.d. non-normal random variables $\{X_i\},\;\; i=1,...,n$, with mean $\mu$ and variance $\sigma^2$. Set the sample mean and the sample variance as  $$\bar x = \frac 1n \sum_{i=1}^nX_i,\;\;\; s^2 = \frac 1{n-1} \sum_{i=1}^n(X_i-\bar x)^2$$  We know that  $$E(s^2) = \sigma^2, \;\;\; \operatorname {Var}(s^2) = \frac{1}{n} \left(\mu_4 - \frac{n-3}{n-1}\sigma^4\right)$$  where $\mu_4 = E(X_i -\mu)^4$, and we restrict our attention to distributions for which what moments need to exist and be finite, do exist and are finite.  Does it hold that  $$\sqrt n(s^2 - \sigma^2) \rightarrow_d N\left(0,\mu_4 - \sigma^4\right)\;\; ?$$
24398
1) Up until now, during practical work sessions, I always used these formulas for uncertainty propagation:  if $C = A+B$  or  $C = A-B$ $$\Delta C = \Delta A + \Delta B$$ if $C = AB$ or $C = \frac{A}{B}$ $$\frac{\Delta C}{C} = \frac{\Delta A}{A}  + \frac{\Delta B}{B}$$ if $C = A^m$ $$\frac{\Delta C}{C} = |m|\frac{\Delta A}{A}$$   These formulas are derived from the expression for the differential of a function of multiple variables:  if $C = f(A,B)$ $$ dC = \frac{\partial C}{\partial A}dA + \frac{\partial C}{\partial B}dB \Rightarrow \Delta C = \frac{\partial C}{\partial A}\Delta A + \frac{\partial C}{\partial B}\Delta B$$  and I think that makes sense because the value C is just a function of two variables that happen to be measurements.  2) But this morning, I was told that this is wrong and I should actually use these instead:  if $C = A+B$  or  $C = A-B$ $$(\Delta C)^2 = (\Delta A)^2 + (\Delta B)^2$$ if $C = AB$ or $C = \frac{A}{B}$ $$\left(\frac{\Delta C}{C}\right)^2 = \left(\frac{\Delta A}{A}\right)^2  + \left(\frac{\Delta B}{B}\right)^2$$  which are apparently derived from a general formula:  $$(\Delta C)^2  = \left(\frac{\partial C}{\partial x_1}\Delta x_1\right)^2 + \left(\frac{\partial C}{\partial x_2}\Delta x_2\right)^2 +   \hspace{0.3cm}...$$  or   $$\Delta C  = \sqrt{\left(\frac{\partial C}{\partial x_1}\Delta x_1\right)^2 + \left(\frac{\partial C}{\partial x_2}\Delta x_2\right)^2 +   \hspace{0.3cm}...}$$  3) So:   Why is this formula better?  Where does it come from? What does it actually represent? (Do I recognize the shape of a norm in that last formula?)  What is wrong with the formula of the differential?   I'm asking a lot, but the propagation of uncertainty always confused the ship out of me.
53337
I was trying to derive equation (65) in the following review:    This slightly unusual then usual classical mechanics because it includes a variation of time also, $\delta t$.  Usually one would define, $$\delta S~=~ \int \left[ L \left(\tilde{q}(t),\dot{\tilde{q}}(t),t\right) -  L \left(q(t),\dot{q}(t),t\right)\right] dt , $$ where,  $$\tilde{q}~=~q+\delta q. $$  We have then (before applying int by parts),  $$ \delta S ~=~ \int \left(\frac{\partial L}{\partial q}\delta q +\frac{\partial L}{\partial \dot q}\delta \dot q\right)dt.$$  How does one proceed if both $q$ and $t$ vary and further that $q$ depends on a varying $t$?  Is the definition now  $$\delta S~=~ \int \left[ L \left(\tilde{q}(t+\delta t),\dot{\tilde{q}}( t+\delta t), t+\delta t\right) -  L \left(q(t),\dot{q}(t),t\right)\right] dt~?  $$  If so how does one proceed?
76134
I got a problem of calculating $E[e^X]$, where X follows a normal distribution $N(\mu, \sigma^2)$ of mean $\mu$ and standard deviation $\sigma$.  I still got no clue how to solve it. Assume $Y=e^X$. Trying to calculate this value directly by substitution $f(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\, e^{\frac{-(x-\mu)^2}{2\sigma^2}}$ then find $g(y)$ of $Y$ is a nightmare (and I don't know how to calculate this integral to be honest).  Another way is to find the inverse function. Assume $Y=\phi(X)$, if $\phi$ is differentiable, monotonic, and have inverse function $X=\psi(Y)$ then $g(y)$ (PDF of random variable $Y$) is as follows: $g(y)=f[\psi(y)]|\psi'(y)|$.  I think we don't need to find PDF of $Y$ explicitly to find $E[Y]$. This seems to be a classic problem. Anyone can help?
65010
Let $(B_t)_{t\geq 0}$ be a standard Brownian motion in $\mathbb{R}^d$. It is intuitive that, for fixed $s&lt;t&lt;u$  $$\mathbb{E}[B_t\mid \sigma(B_s,B_u)]=B_s+\frac{t-s}{u-s}(B_u-B_s).$$  However, I cannot think of a way to show this rigorously.   If first attempted to take $A\in\sigma(B_s,B_u)$ and show that $\mathbb{E}[1_A B_t]=\mathbb{E}[1_A(B_s+\frac{t-s}{u-s}(B_u-B_s))]$. But I cannot manage to show this equality.   I'd be very thankful for any ideas and suggestions on how to tackle this problem.
98645
Let us consider a (quantum) Harmonic oscillator: $$H=\frac{p^2}{2m}+\frac{1}{2} m \omega^2 x^2$$ Using the annihilation/creation operators defined as: $$a=\sqrt{\frac{\hbar}{2 m \omega}}(x+\frac{i}{m \omega}p)$$ $$a^{\dagger}=\sqrt{\frac{\hbar}{2 m \omega}}(x-\frac{i}{m \omega}p)$$ I end up with $H=\hbar \omega a^{\dagger} a$ Now, let's assume that for any reason I know that the dynamic should be restricted to the two first levels only. One way that I should we could do is to consider: $a \to \sigma_-, a^{\dagger} \to \sigma_+$ Then, we find: $$H=\hbar \omega \sigma_+ \sigma_-=\hbar \omega |1\rangle \langle 1|$$ But what confuses me is that if I do the replacement in $p$ and $x$, I don't find the same. Indeed we have: $$p=i\sqrt{\frac{\hbar m \omega}{2}}(a^{\dagger}-a)$$ $$x=\sqrt{\frac{\hbar}{2 m \omega}}(a^{\dagger}+a)$$ Performing the replacement there, I find: $$p=\sqrt{\frac{\hbar m \omega}{2}}\sigma_y$$ $$x=\sqrt{\frac{\hbar}{2 m \omega}}\sigma_x$$ And as $\sigma_y^2=\sigma_x^2=I$, I would end up with $H \propto I$ from the first Hamiltonian I wrote. Another annoying example. Imagine that my oscillator as in addition an interacting term of the form: $$H_{\text{int}}=p^2*F(t)$$ Where $F(t)$ is a driving field for instance. With the two level approximation and the mapping considered, as $p^2 \propto I$, the interaction would disappear. What is the procedure to cut to its two first level an Harmonic oscillator to get a qubit. I feel like I don't understand anymore. Typically to what the momentum and position should be mapped to ? How to solve the &quot;paradox&quot; I get to ?
93945
I'm currently stuck on a problem where I have to integrate on a particular set defined through a dirac delta function. If I understood correctly it all boils down to using the curved analogous of  $$  \int_{\mathbb{R}^n} f(\vec{x}) \delta(\phi(\vec{x}))|\nabla\phi|d^nx \tag{1} $$  my guess is that in GR  this expression becomes either  $$ \int_\Omega f(x) \delta(\phi(x))\sqrt{g^{\mu\nu}\frac{\partial \phi(x)}{\partial x^\mu}\frac{\partial \phi(x)}{\partial x^\nu}} \sqrt{ -g}\,d^4 x \tag{2}$$  or  $$ \int_\Omega f(x) \delta(\phi(x))\sqrt{g^{\mu\nu}\frac{\partial \phi(x)}{\partial x^\mu}\frac{\partial \phi(x)}{\partial x^\nu}}\,d^4 x \tag{3}$$  given that in GR $$\delta^4 (\xi) =\frac{\delta^4(x)}{\sqrt{-g}}\tag{4}$$ where the $\xi$ are flat coordinates, but here I have a $\delta^{1}$, not a $  \delta^{4}! $ So I'm thinking the first formula is the correct one. Am I right?  Which one is correct?  PS. It would be awesome if someone had any sort of useful reference on the topic!
66693
I was reading this  where it had a gaussian distribution model.   I mean gaussian is given by  $P(y) = \frac{e^{-\frac{1}{2}(y -\mu)^T \Sigma^{-1}(y -\mu)}}{2\pi^{n/2}|\Sigma|^{1/2}}$  But is   $\frac{e^{-(y -\mu)^T \Sigma^{-1}(y -\mu)}}{2\pi^{n}|\Sigma|}$ Gaussian as well?  The paper used their model saying the gaussian had the above form. It is valid?
10814
I have a number of measurements of the same quantity (in this case, the speed of sound in a material). Each of these measurements has their own uncertainty.  $$ v_{1} \pm \Delta v_{1} $$ $$ v_{2} \pm \Delta v_{2} $$ $$ v_{3} \pm \Delta v_{3} $$ $$ \vdots $$ $$ v_{N} \pm \Delta v_{N} $$  Since they're measurements of the same quantity, all the values of $v$ are roughly equal. I can, of course, calculate the mean:  $$ v = \frac{\sum_{i=1}^N v_{i}}{N}$$  What would the uncertainty in $v$ be? In the limit that all the $\Delta v_i$ are small, then $\Delta v$ should be the standard deviation of the $v_i$. If the $\Delta v_i$ are large, then $\Delta v$ should be something like $\sqrt{\frac{\sum_i \Delta v_i^2}{N}}$, right?  So what is the formula for combining these uncertainties? I don't think it's the one given in  (though I may be wrong) because it doesn't look like it behaves like I'd expect in the above limits (specifically, if the $\Delta v_i$ are zero then that formula gives $\Delta v = 0$, not the standard deviation of the $v_i$).
96591
Given that a random variable $X$ is a normally distributed variable with mean $\mu$ and variance $\sigma^2$. Why exactly is:  $$\mathbb{E}[|X|] = \mathbb{E}[X\mid X&gt;0]P(X&gt;0) - \mathbb{E}[X\mid X&lt;0]P(X&lt;0)$$  Because we have that:  $$\mathbb{E}[X\mid X&gt;0] = \frac{\int_0^\infty xf_X(x)\,dx}{P(X&gt;0)}$$  Therefore my definition immediately above must be incorrect?  Because if it is not we have that:  $$\mathbb{E}[|X|] = \int_0^\infty xf_X(x)\,dx - \int_{-\infty}^0 xf_X(x)\,dx$$  and where does the minus there come from anyway, and why?
110892
Since integration is not my strong suit I need some feedback on this, please:  Let $Y$ be $\mathcal{N}(\mu,\sigma^2)$, the normal distrubution with parameters $\mu$ and $\sigma^2$. I know $\mu$ is the expectation value and $\sigma$ is the variance of $Y$.  I want to calculate the $n$-th central moments of $Y$.  The density function of $Y$ is $$f(x)=\frac{1}{\sigma\sqrt {2\pi}}e^{-\frac{1}{2}\left(\frac{y-\mu}{\sigma}\right)^2}$$  The $n$-th central moment of $Y$ is $$E[(Y-E(Y))^n]$$  The $n$-th moment of $Y$ is $$E(Y^n)=\psi^{(n)}(0)$$ where $\psi$ is the Moment-generating function $$\psi(t)=E(e^{tX})$$  So I started calculating:  $$\begin{align}  E[(Y-E(Y))^n]&amp;=\int_\mathbb{R}\left(f(x)-\int_\mathbb{R}f(x)dx\right)^n\,dx \\ &amp;=\int_\mathbb{R}\sum_{k=0}^n\left[\binom{n}{k}(f(x))^k\left(-\int_\mathbb{R}f(x)dx\right)^{n-k}\right]\,dx \\ &amp;=\sum_{k=0}^n\binom{n}{k}\left(\int_\mathbb{R}\left[(f(x))^k\left(-\int_\mathbb{R}f(x)dx\right)^{n-k}\right]\,dx\right) \\ &amp;=\sum_{k=0}^n\binom{n}{k}\left(\int_\mathbb{R}\left[(f(x))^k\left(-\mu\right)^{n-k}\right]\,dx\right) \\ &amp;=\sum_{k=0}^n\binom{n}{k}\left((-\mu)^{n-k}\int_\mathbb{R}(f(x))^k\,dx\right) \\ &amp;=\sum_{k=0}^n\binom{n}{k}\left((-\mu)^{n-k}E\left(Y^k\right)\right) \\ \end{align}$$  Am I on the right track or completely misguided? If I have made no mistakes so far, I would be glad to get some inspiration because I am stuck here. Thanks!
7910
It is known that for a system at  described by the , the probability of being in a state of energy $E$ at temperature $T$ is given by the : $$ \tag{1} P(\text{state with energy } E) \propto e^{-E/kT}. $$  I have no problems with the mathematical derivation of this result, but I do wonder if there is a way to understand what (1) means intuitively.  In other words: Is there a way to "guess" that the correct distribution for such a system is given by (1), using solely physical arguments?
93153
Please forgive my lack of maths knowledge,  It is my understanding that:  Standard Deviation is the average distance from the mean in a data set of numbers.  Therefore it stands to reason that working out the standard deviation of the data set $x_i = \{1,2,3,4,5\}$ would involve the following.  First working out the mean $\mu(x_i) = 3$ and Then working out the sum of the distance from the average $\sum{|x_i-\mu|} = 6$ then we do $\frac{\sum{|x_i-\mu|}}{N} = \frac{6}{5} = 1.2$  This means that, according to my method/thinking, 1.2 is the standard deviation.  However when using the formula $\sqrt{\frac{\sum{(x_i-\mu)}^2}{N}}$ I get $1.414$  Can someone explain why I'm wrong in layman terms. Thankyou
13181
Just beginning to learn quantum mechanics and my textbook says that all pure eigenstates of the Schrodinger equation ( with a specified potential function) have a zero standard deviation for the expectation values of energy.So every measurement of the energy yields a fixed value. I've also read that pure eigenstates are physically realisable solutions in some situations ( like the infinite potential well). But wouldn't that violate the uncertainty principle? the standard deviation in expectation of energy is zero and hence $$ΔE~Δt=0.$$ Where am I going wrong? Also could someone also give me an insight as to what uncertainty in time even means? I haven't been able to grasp that concept clearly yet.
88039
Today I taught an introductory class of statistics and a student came up to me with a question, which I rephrase here as: "Why is the standard deviation defined as sqrt of variance and not as the sqrt of sum of squares over N?"  We define population variance: $\sigma^2=\frac{1}{N}\sum{(x_i-\mu)^2}$   And standard deviation: $\sigma=\sqrt{\sigma^2}=\frac{1}{\sqrt{N}}\sqrt{\sum{(x_i-\mu)^2}}$.   The interpretation we may give to $\sigma$ is that it gives the average deviation of units in the population from the population mean of $X$.   However, in the definition of the s.d. we divide the sqrt of the sum of squares through $\sqrt{N}$. The question the student raises is why we do not divide the sqrt of the sume of squares by $N$ instead. Thus we come to competing formula: $$\sigma_{new}=\frac{1}{N}\sqrt{\sum{(x_i-\mu)^2}}.$$ The student argued that this formula looks more like an "average" deviation from the mean than when dividing through $\sqrt{N}$ as in $\sigma$.  I thought this question is not stupid. I would like to give an answer to the student that goes further than saying that the s.d. is defined as sqrt of the variance which is the average squared deviaton. Put differently, why should the student use the correct formula and not follow her idea?  This question relates to an older thread and answers provided . Answers there go in three directions:   $\sigma$ is the root-mean-squared (RMS) deviation, not the "typical" deviation from the mean (i.e., $\sigma_{new}$). Thus, it is defined differently. It has nice mathematical properties. Furthermore, the sqrt would bring back "units" to their original scale. However, this would also be the case for $\sigma_{new}$, which divides by $N$ instead.   Both of points 1 and 2 are arguments in favour of the s.d. as RMS, but I do not see an argument against the use of $\sigma_{new}$. What would be the good arguments to convince introductory level students of the use of the average RMS distance $\sigma$ from the mean?
74559
I am slightly confused about the application of . Which during standard derivations indicates a probability of transitioning from the state $|i \rangle$ to the state $|f\rangle$ of: $$P=\frac{2\pi t}{\hbar^2} | \langle f | \hat V | i \rangle|^2 \delta(E_f-E_i-\hbar \omega)$$ This holds for large values of $t$ but for arbitrarily large values of $t$: $$P&gt;1$$ Which (to me anyway) makes no physical sense since there can't be a probability greater then 1 of been in the final state. If my interpretation is right then how is this problem resolved/allowable? and if it is wrong what part of what I have said is wrong?