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1

I would say the answer is rather simple. If you do various experiments to measure the position (or momentum) of a particle, then QM only gives you consistent and accurate predictions when the mathematical symbols $x$ and $p$ are associated with the measured position or momentum. However, position and momentum are not the only canonically conjugate operators ...


3

The answer may be deeper than you expect. The operator $K$ that the OP defined --- here it will be called $-i\nabla_x$ --- is the generator of space translations. And the operator $X$ --- here called $x$ --- is the generator of momentum translations. If you exponentiate them, to form the corresponding one-parameter unitary groups of translation, you get ...


4

In quantum systems it is not possible to perform a measurement without affecting the measured system. This is because, roughly speaking, the interaction with the instrument creates a correlation between the system and the instrument whose effective result is a modification of the system's state. Therefore if you perform repeated measurements of the same ...


1

Young's modulus of wire material is given by, \begin{align} Y=\frac{\text{stress}}{\text{strain}}=\frac{F/A}{l/L}=\frac{4FL}{\pi d^2 l}. \end{align} From given data, $F=mg=9.8\;\mathrm{N}$, $L=2.0\;\mathrm{m}$, $l=0.8\times{10}^{-3}\;\mathrm{m}$, and $d=0.4\times{10}^{-3}\;\mathrm{m}$. Substitute the values in equation to get $Y=1.95\times{10}^{11}\approx ...


2

NOTE: This answer has now been merged into Understanding the quantum eraser from a quantum information stand point (part IV). Let me start by copying the first part of my previous answer which describes the circuit model of a double-slit or other interference experiment; then, I will try to describe the delayed choice setting (the way I understand it). ...


6

The answer is structured as follows: I will first give the quantum circuit corresponding to a normal double slit (or interferometer), then the circuit where the which-way information has been recorded, a circuit where the which-way information is first recorded and then erased in a unitary way, and finally a circuit where the which-way information is ...


0

The Heisenberg uncertainty principle establishes a lower bound on the uncertainty (hence the $\lt$ sign). It follows that if you have additional error (due to your measurement instrument) it can only increase the uncertainty over this quantum mechanical limit.


2

Yes, indeed any measurement where you learn something will disturb the state. This is also known as "no information without disturbance". Recently, there has been a lot of flurry of activity trying to pin this down quantitatively. As you said, this is not what the Heisenberg relation you write down is about, but the effect still certainly exists. Here is a ...


4

Actually while measuring, the Probability Distribution function of a particle also changes, Does this means that the measuring instrument has some effect ? The measuring process may change the boundary conditions of the solutions of the quantum mechanical equations of the system under measurement, so the complex conjugate square of the wave function ( ...


0

These are standard deviations of a probability distribution indeed. The probability distribution is that of getting a particular value while the system is in the state prior to measurement. So we imagine measuring a state - it collapses to some value, and then somehow resetting time back before measurement and measuring again. If we repeat this process, ...



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