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The generalized uncertainty principle relating two operators $A$ and $B$ is $$\sigma_A^2 \sigma_B^2 \geq {\left( \frac{1}{2i} \langle\left[ A,B\right] \rangle\right)} ^2$$ where $[A,B]=AB-BA$ is the commutator of the operators. This relation was derived using two inequalities. The first is the Schwarz inequality which, in bra- ket notation is ...

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One of the main reasons the virtual particles are used is that in many contexts we do not have a non-perturbative formulation of quantum field theory. What we can do is compute some amplitudes perturbatively (e.g. for outcomes of particle collisions) using Feynman diagrams. These diagrams have input/output lines in them, usually identified with colliding ...

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I henceforth assume $\hbar =1$. There is no reason to introduce Dirac deltas here, everything is elementary. Moreover as the function $\psi$ is not differentiable, one cannot use the form of the momentum operator $P$ as derivative which is valid only on smooth functions. Forcing this way would introduce unnecessary difficulties as the derivative must be ...

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That's the covariance of X and Y. It tells you, in a way, "how much" X and Y are correlated, with covariance 0 meaning uncorrelated (Not to be confused with independent). Because $\langle A\rangle$ and $\langle B\rangle$ can be quite large, it is customary to define the correlation coefficient: ($cov(A,B)$ is the covariance of A and B) ...

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I) One problem is that the momentum operator $\hat{p}$ is an unbounded operator, which means that it is only defined on a domain $D(\hat{p}) \subsetneq {\cal H}$ of the Hilbert space ${\cal H}=L^2(\mathbb{R})$. When we apply the differentiation operator $\hat{p}=\frac{\hbar}{i}\frac{d}{dx}$ to OP's wave function \tag{1} \psi(x)~=~A(a-x)\theta(a-|x|), ...

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Well, you can conclude that something is wrong by the following logic: momentum is an observable, which means its allowed values must be things that you could read off a measuring device (assuming you had one that measures momentum). These are necessarily real values, and since the expectation value is some linear combination of possible measurements, it ...

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The wavefunction has a discontinuity at $x=-a$, which gives a term $-2aA i \hbar \delta(x+a)$ when you act with $p$. The contribution from this to the expectation value of momentum exactly cancels the imaginary value you have calculated. Two more-general points: The momentum operator is hermitian, which means its expectation value must be real (provided ...

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Note that $\Delta p_x \Delta r$ does not satisfy the uncertainty principle in the strict sense since $r$ is not conjugate to $p_x$ (or $p_y$ and $p_z$). Instead you can consider $\Delta p_x \Delta x$. The ground state of the hydrogen atom is $$\psi_0(r) = \frac{1}{\sqrt{\pi a^3}} e^{-r/a},$$ where $a$ is the Bohr radius. First of ...

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No. The Uncertainty Principle has to do with the act of measuring. Basically, you cannot simultaneously measure both position and momentum to an arbitrary degree of accuracy. The more accurately you meausre one, the less accurate your measurement of the other becomes. The uncertainty in momentum , as far as I know, won't result from your not knowing when ...

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It is perfectly possible to use wavelets to analyse quantum mechanical situations. The wavelets are localised in both time and frequency but they are themselves subject to the uncertainty principle - if you want a better time resolution, you need to pay for it with a coarser frequency resolution. The uncertainty principle is a universal wave phenomenon and ...

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As ACuriousMind says in a comment, this isn't the approach Yukawa used in his 1935 paper (Yukawa H 1935 Proc. Phys. Math. Soc. Japan 17 48) though whether he did that calculation in the privacy of his own notebook only he knows. The calculation you describe is a rather arm waving sort of justification for the relationship between the mass of the mediating ...

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I agree with the first half of your question, but I think you have wandered astray in the second part. The theory of quantum mechanics is based on a number of axioms. A thorough description of these can be found in this paper, though this is a rather greater level of detail than would be useful for most non-physicists. If you construct the theory of quantum ...

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My understanding is that the Uncertainty principle or relation is a mathematical consequence of the axioms of quantum mechanics. So in it is an inevitable consequence of the mathematical theory known as quantum mechanics. What can occur in another universe is a irrelevant to any questions of what can occur here. Uncertainty principle is only a physical ...

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In order to measure an object's speed, you need at least two measurements of its position at different times. This is not the case. The radar guns used by police to determine if you are exceeding the speed limit do not use position measurements. They instead measure the frequency difference between the outgoing and reflected signals. No position ...

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The uncertainty principle doesn't say anything about simultaneous measurements of a particle - that's just a myth which originated from Heisenberg's interpretation of it. Let us first describe the basis of quantum physics and let's start with the most innocent looking object: the quantum state. We can see a quantum state as a prescription to prepare a ...

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The uncertainty principle says something a deeper than "it is impossible to measure both position and momentum to arbitrary accuracy". It says 1) The accuracy is precisely limited by $\Delta x \Delta p > \hbar/2$. 2) In fact, this is not a limit of our measuring procedure, but a limit of reality. If something has well-defined position, it does not have ...

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I think the question is correct. While i was reading the bra ket algebra over and over again i wonder where the hell these are just the properties of quantum mechanics. Following the derivation of uncertanity relation i have found that all the arguments can be applied to classical mechanics equally well. Except for one postulate that after the measurement ...

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