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72 votes

Why is it impossible to measure position and momentum at the same time with arbitrary precision?

When someone asks "Is it really impossible to simultaneously measure position and momentum with arbitrary precision in quantum theory?", the best preliminary answer one can give is another ...
pglpm's user avatar
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40 votes
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The importance of the phase in quantum mechanics

When people say that the phase doesn't matter, they mean the overall, "global" phase. In other words, the state $|0 \rangle$ is equivalent to $e^{i \theta} |0 \rangle$, the state $|1\rangle$ is ...
knzhou's user avatar
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36 votes
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Is my interpretation of the underlying idea behind this 2022 Nobel Prize story “How physicists proved the universe isn’t real” more or less accurate?

First, it's important to clarify two definitions. The principle of locality loosely says that interactions occur at individual points in space. If you say that one charged particle at a point $\...
J. Murray's user avatar
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35 votes

Am I in a superposition?

This is a deceptively simple question, and I would like to answer it in the context of Condensed Matter Theory. We will start with the question: why do objects in our everyday life appear to be ...
Shiki Ryougi's user avatar
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31 votes
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Why do eigenvalues correspond to observable quantities?

Suppose we don't know quantum mechanics yet and we want to calculate the expecatation value of an observable $A$. Could be momentum, spin whatever. It is given by $$\mathbb E(A)=\sum_i a_i\,p(a_i)$$ ...
AccidentalTaylorExpansion's user avatar
27 votes
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Why are quantum effects of the apparatus ignored in quantum experiments?

And in all the experiments all these in-between steps are considered to be "magically perfect". Nobody pays them any more attention except to mention that they're there. All mirrors reflect ...
AwkwardWhale's user avatar
23 votes
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Why are wavefunction collapses instantaneous?

It’s important to remember that quantum mechanics is a tool that we use to describe the world — it is not the same as the world. For all that we love to talk about wavefunctions, it’s not clear at ...
rob's user avatar
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23 votes
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Why are expectation values of an observable important in QM?

This is an important point which should be discussed. What one obtains from experiments are frequencies of outcomes of given measured observables on an ensemble of identical quantum systems all ...
Valter Moretti's user avatar
21 votes
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Why do we use Eigenvalues to represent Observed Values in Quantum Mechanics?

One of the postulates of quantum mechanics is that for every observable A, there corresponds a linear Hermitian operator A^, and when we measure the observable A, we get an eigenvalue of A^ as the ...
J. Murray's user avatar
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20 votes

Are observables in QFT actually observable?

A quantum mechanical secret kept very well in plain sight is that the word "observable" when defined as "self-adjoint operator on a Hilbert space" does not actually mean "you ...
ACuriousMind's user avatar
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20 votes
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Does the collapse of the wave function depend on the observer?

The measurement problem is one of the most relevant open problems of quantum mechanics. What is a measurement? What constitutes an observer and what doesn't? Is the wavefunction a physical object (...
Prallax's user avatar
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19 votes

Do we or do we not observe (measure) superpositions all the time?

To answer the question, you have to specify what observable’s eigenstates the “superpositions” you’re talking about are superpositions of. If you measure the energy, you always get one single energy ...
G. Smith's user avatar
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18 votes
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Why is it impossible to measure position and momentum at the same time with arbitrary precision?

You can't measure precise values at the same time because precise values for both don't exist at the same time. All the properties of, say, an electron and be inferred from the electron's wave ...
mmesser314's user avatar
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17 votes
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Confusion about the description of the uncertainty principle

The first one is correct, the second is not. The second definition1 is actually describing the observer effect. Explanations written by non-experts often mix the two up. But one key difference is ...
David Z's user avatar
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17 votes
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Are two states with the same measurement probabilities necessarily equal up to unitary equivalence?

Let $H$ denote a finite-dimensional complex Hilbert space, $\rho$ and $\rho^\prime$ be two density matrices, i.e. positive semi-definite operators with unit trace and $A$ an arbitrary hermitian ...
Tobias Fünke's user avatar
17 votes
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What is the Hamiltonian of a Measurement?

The following argument is very vague and hand-wavey - as is inevitable when discussing measurement apparatus with on the order of $10^{23}$ particles. It is intended only to illustrate the idea. We ...
Nullius in Verba's user avatar
16 votes

The importance of the phase in quantum mechanics

The global phase does not matter. In your example $\lambda(\vert\psi_1\rangle+\vert\psi_2\rangle)$ has the same physical contents as $\vert\psi_1\rangle+\vert\psi_2\rangle$ but this will be in ...
ZeroTheHero's user avatar
15 votes
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Can the collapse of the wave function be modelled as a quantum system on its own?

To model the act of measurement itself as an interaction of the measurement apparatus and the measured system as quantum systems is a perfectly standard thing to do, though you might get disagreements ...
ACuriousMind's user avatar
  • 126k
14 votes

Proof that you can't disentangle two parties if you only operate on one

If there was some unitary operator factorized as $\mathbb I_A \otimes U_B$ that would send the entangled state $|\psi\rangle$ to a factorized state $|\phi_A\rangle\otimes |\phi_B\rangle$, ie : $$|\...
SolubleFish's user avatar
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12 votes

Why do eigenvalues correspond to observable quantities?

I think the conceptual issue is that you've got it backwards. Given any set of eigenfunctions, you can have operators with those eigenfunctions, whose corresponding eigenvalues are anything you want. ...
knzhou's user avatar
  • 103k
12 votes

How is it that a particle's wave function is not a real thing, yet we can still observe it?

Plenty of things in physics are mathematical objects/models that we use to describe the world around us and can also use to guide our measurements and observations. For example, when you are walking ...
BioPhysicist's user avatar
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11 votes

A question regarding the commutators of operators

I think that the misunderstanding follows from the following fact. Given that $[A,B]=0$ then we can build a simultaneous eigenbasis for $A$ and $B$, call it $\mathcal{B}_1$. On the other hand we have ...
Davide Morgante's user avatar
11 votes

Are observables in QFT actually observable?

As emphasized in the excellent answer by ACuriousMind, some of the operators that we usually call "observables" might not be feasibly measurable through any physical process that the theory ...
Chiral Anomaly's user avatar
11 votes

Am I in a superposition?

TL;DR Your position is not well-defined... but for more mundane reasons. Quantum mechanical view You are not in a pure state - that is you are not an object, that can be described by a wave function, ...
Roger V.'s user avatar
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11 votes
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Which experiments confirm that the Born rule is $|\psi|^2$ rather than $|\psi|$?

I might be missing the depth somewhere, but I would suggest the electron double-slit experiment first: $|\sin(kx)|$ clearly looks different as compared to $\sin^2(kx)$. The second example is the Malus'...
John's user avatar
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11 votes
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What is the physical significance of projectors vs. Pauli matrices in spin measurements?

The projection operator $|u\rangle\langle u|$ corresponds to the experimental question "Does the spin point in the positive $z$-direction - yes or no?" The outcome "yes" ...
Hyperon's user avatar
  • 6,676
10 votes

Why do we use Eigenvalues to represent Observed Values in Quantum Mechanics?

The Stern-Gerlach experiment and similar experiments show A system has a state. The states of a system form a Hilbert space. You can choose a set of basis states and represent the system's current ...
mmesser314's user avatar
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10 votes

Am I in a superposition?

The question 'where am I', taken at a microscopic level, is hardly meaningful. A human body, as a physical system, is overwhelmingly complex. It certainly cannot be described as a quantum system in ...
Stéphane Rollandin's user avatar
10 votes

Are two states with the same measurement probabilities necessarily equal up to unitary equivalence?

Actually $\rho=\rho'$. Indeed, specializing $A= P= |\psi\rangle \langle \psi|$ for $||\psi||=1$, the hypothesis implies $\langle\psi| (\rho-\rho')\psi\rangle =0$. Linearity permits to relax the ...
Valter Moretti's user avatar
10 votes

What reason/evidence do we have to think that the Planck length is the smallest length possible?

Or are there only amateurs who think so? Bingo :) The Planck units don’t give sharp limits on the existence of quantities like length, time, etc (or rather, there’s no reason to think that they do). ...
J. Murray's user avatar
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