# Why is the position of a massive particle not blurred?

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• fxdung
In summary: For the state above to be an equal superposition of both macro states it would have to have a complex number with a norm of 1 in front of it, i.e. the state would have to be:$$\frac{1}{\sqrt{2}}\left(|D_{\uparrow}\rangle + e^{i\theta}|D_{\downarrow}\rangle\right)$$which would have a phase in front of the second term. This phase can be absorbed into the phase of ##|D_{\downarrow}\rangle## so the state would actually be:$$|D_{\uparrow fxdung In QM a free micro particle with definite momentum has not definite positions.But why does in classical physics a free particle with definite momentum still have definite positions but not blurred as in QM?(Because we can deduce Classical Physics from QM) Classical physics breaks down at the subatomic level. Macroscopic objects with a definite momentum still have a blurred position but the blurring is so infinitesimal that classical physics is good enough. Heisenberg's uncertainty principle, expressed as the product of position and momentum standard deviations being greater than a constant, i.e. ##\sigma_x \sigma_p \geq \frac{\hbar}{2}##, apply to macro objects also. However, the uncertainty of the position and momentum of a macro object compared to its size is so tiny that the uncertainty principle is safe to ignore. Not so with subatomic particles. fxdung said: why does in classical physics a free particle with definite momentum still have definite positions but not blurred as in QM?(Because we can deduce Classical Physics from QM) Because classical physics is the limit ##\hbar\to 0##. In this limit the uncertainty relation becomes vacuous. The limit ##\hbar \rightarrow 0## doesn't make physical sense. It's only a formal idea to study the semi-classical limit, e.g., by applying singular perturbation theory to the Schrödinger equation, known as the WKB method. The physics of the limit becomes more clear by using the equivalent path-integral approach: There it becomes clear that the classical limit in this sense becomes valid when the situation is such that the action around the classical particle trajectory, such that the exponential ##\exp(\mathrm{i} S[x]/\hbar)## is rapidly oscillating around the classical trajectory except at the classical trajectory itself, because ##|S[x]|/\hbar \gg 1##. This makes the saddle-point approximation of the path integral, which leads to the WKB approximation, becomes valid. fxdung said: In QM a free micro particle with definite momentum has not definite positions.But why does in classical physics a free particle with definite momentum still have definite positions but not blurred as in QM?(Because we can deduce Classical Physics from QM) The above responses are not wrong, they simply have expressed the idea in different ways. I would express it differently again and say even for macro objects they neither have 100% definite position nor momentum - it's better modeled as a wave packet but the packet size is so small it likely could not even be measured with current technology. See Gaussian wave packets in quantum mechanics: https://en.wikipedia.org/wiki/Wave_packet As noted in the article wave packets tend to spread - so why don't macro wave-packets spread and eventually have no definite position. The answer is decoherence - even a few stray photons from the CBMR will give a dust particle pretty close to definite position. So what is really going on is its interaction with the environment is part of a balancing act of the wave-packed spreading and the interaction keeping the spreading in check. Now you are probably asking why does decoherence tend to give definite positions - its to do with the radial nature of most interactions. The detail can be found in:: https://www.amazon.com/dp/3540357734/?tag=pfamazon01-20 I need to mention not to get too worried if people explain things in different ways - that happens a lot around here - while it can be confusing at first the different perspectives will hopefully deepen understanding and perhaps lead to more questions that clarify things even further. BTW the best way to deduce classical physics from QM is by Feynman's (actually it was Dirac who first discovered it - nice little bit of trivia to look into) path integral approach which leads to the principle of least action. Once you have that then you have classical mechanics because that and symmetry is pretty much all that is needed. For the details see Landau - Mechanics. As testament to how good that book is where have you seen reviews like this before: https://www.amazon.com/dp/B00BCRB50E/?tag=pfamazon01-20 Now you know the full breath of modern physics. A lot of it is the principles of QM and symmetry - even classical physics. When the early quantum founders let the genie out of the bottle they started a revolution that nowadays is as close as we have to a theory of everything. Thanks Bill DarMM and andrew s 1905 fxdung said: In QM a free micro particle with definite momentum has not definite positions.But why does in classical physics a free particle with definite momentum still have definite positions but not blurred as in QM?(Because we can deduce Classical Physics from QM) It is closely related to the law of large numbers in the theory of probability. If you flip one coin, you cannot predict whether you will get one heads or one tails. This inability to predict the result of flipping the coin makes the coin blurred. But if you flip billion coins, it is almost certain that about half of the coins will be heads and about half of them will be tails. With billion coins you are almost certain what will happen, so it's not blurred much. This explains why big systems made of many blurred objects look as if they are not blurred. That's why classical objects, which are objects made of many atoms, don't look blurred. DarMM, bhobba and vanhees71 And it's also because you look at macroscopic objects already only in a blurred way. The location of a macroscopic object is almost never needed to such a degree of accuracy that quantum effects become important. The only practical example that comes into my mind right now for an exception are the mirrors of the LIGO gravitational-wave observatory, where quantum noise is already an issue as far as I know. DarMM, Demystifier and bhobba The classical limit is quite complex. Decoherence shows us that as systems get larger and let's say we decompose the system into its micro and macro degrees of freedom:$$\mathcal{H}_{total} = \mathcal{H}_{macro}\otimes\mathcal{H}_{micro}$$then interference effects become incredibly small in ##\mathcal{H}_{macro}## and superpositions between macroscopic degrees of freedom represent only classical ignorance. So if we have a device that can be in one of two states ##|D_{\uparrow}\rangle## and ##|D_{\downarrow}\rangle##, then states like:$$|D_{\uparrow}\rangle + e^{i\theta}|D_{\uparrow}\rangle$$will always have terms with ##\theta## be ##\approx 0## and observables which can detect ##\theta## are not physically realizable. Thus not only are terms containing ##\theta## small, but all observables you can actually measure commute so you can't access ##\theta## anyway. The last recourse might be to detect correlations between micro and macro components, effectively entanglement. However observables that can detect entanglement become similarly rare and impossible to construct. EDIT: Altered from @vanhees71 's post below. Last edited: bhobba I'd rather argue a bit differently here. A general pure state is given by$$\hat{\rho}(\theta)=|\psi \rangle \langle \psi|$$with$$\psi =\frac{1}{\sqrt{2}} (|D_{\uparrow} \rangle+\exp(\mathrm{i} \theta) |D_{\downarrow} \rangle).$$Then through the interaction with the many degrees of freedom making up the macroscopic body, the relative phase ##\theta## is rapidly changing with time. Averaged over macroscopic small but microscopic large times you find as an effective statistical operator$$\hat{\rho}'=\langle {\hat{\rho}(\theta)} \rangle_{\theta}=\frac{1}{2} \hat{1}.
Nothing in any way prefers ##\theta=0## here, but via "coarse-graining" through the interactions between the microstate (single spin in this example) with a microscopic system the phase gets completely randomized (and FAPP uncontrollable), leading to an effective "thermalization" of the state of the microscopic system, i.e., the maximum-entropy state, with neither spin-orientation nor any phase information left.

Mentz114, weirdoguy and DarMM
You're right of course. I should say off-diagonal terms in the density matrix containing ##\theta## are driven to zero, not that ##\theta## itself is.

vanhees71

## 1. Why is the position of a massive particle not blurred?

The position of a massive particle is not blurred because it follows the laws of classical mechanics, which state that the position and momentum of a particle can be precisely determined at any given time.

## 2. How does the mass of a particle affect its position?

The mass of a particle does not directly affect its position. However, the mass does affect the particle's momentum, which in turn affects its position according to the laws of motion.

## 3. Can the position of a massive particle ever be completely precise?

No, according to the Heisenberg uncertainty principle, it is impossible to know the exact position and momentum of a particle simultaneously. There will always be a degree of uncertainty in the measurement of a particle's position.

## 4. Why is the position of a massive particle more precise than that of a subatomic particle?

The position of a massive particle is more precise because it has a larger mass and therefore a smaller wavelength, making it easier to determine its position with greater accuracy.

## 5. How does the concept of wave-particle duality relate to the position of a massive particle?

The concept of wave-particle duality states that particles can exhibit both wave-like and particle-like behavior. This means that while the position of a massive particle may appear to be precise, it is still subject to the uncertainty principle and can exhibit wave-like properties in certain situations.

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