71 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 in quantum theory?", the best preliminary answer one can give is another question: "what do ...
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39 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 ...
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  • 95.4k
24 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 $$\mathbb E(A)=\sum_i a_i\,p(a_i)$$ ...
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21 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 ...
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  • 71.9k
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 ...
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19 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 ...
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  • 52k
19 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 ...
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  • 107k
18 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 (...
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  • 2,627
17 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 ...
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17 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 ...
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  • 27.3k
16 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 ...
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  • 73.9k
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 : $$|\...
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  • 4,026
13 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. ...
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13 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 ...
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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 ...
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  • 4,267
11 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 ...
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  • 53.4k
10 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 ...
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10 votes
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Measurement on mixed states

As mentioned by the OP both versions are the same. For an observable $A$ of the form $$A = \sum\limits_k a_k \, P_k \quad , $$ with the projections $P_k^2 =P_k = P_k^\dagger$ on the eigenspace ...
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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 ...
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9 votes

The importance of the phase in quantum mechanics

While the other answers are correct, this is not a different answer but rather an illustration that the relative phase is indeed important in quantum mechanics. We know that bosons (particles with ...
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  • 247
9 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 ...
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  • 27.3k
8 votes

Is information conserved in quantum mechanics (after wave function collapse)?

Short answer: the collapse of a wavefunction destroys information. As you correctly said, as long as the quantum state evolves according to the Schrodinger equation, information is conserved. If we ...
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  • 1,564
8 votes

Do quantum measurements violate conservation laws?

No. Remember that, even before quantum mechanics, conservation laws apply only to closed systems. When a "quantum measurement" is performed, the previously-closed system becomes an open ...
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7 votes
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How is conditional probability handled in quantum mechanics?

Yes, conditional probabilities are studied in quantum mechanics, especially from a quantum informational perspective. The trickiest/interesting aspect of it is that, because of the nature of QM, it ...
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  • 12.4k
7 votes
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What is the probability of obtaining the same measurement a finite time after causing a wave-function collapse with my initial measurement?

This is a problem of time evolution of a state. If your initial state is $|\psi\rangle(t=0)$, then the state at later time $t$ is given by, $$|\psi\rangle(t)=e^{\frac{i\hat{H}t}{\hbar}}|\psi\rangle(0)$...
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  • 914
7 votes

A question regarding the commutators of operators

A mistake made here is this statement and all similar ones: But since $[A,B]=0$ we can measure $B$ exactly subsequent to $A$ because they are in the same eigenstates. The collapse to an eigenstate ...
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  • 601
7 votes
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Is the Copenhagen interpretation still the most widely accepted position?

This is very hard to quantify, particularly because it's hard to define the population that should be under consideration. The best(-known) attempt to do this is reported in the paper A Snapshot of ...
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7 votes

Is the Copenhagen interpretation still the most widely accepted position?

It's difficult to answer this question because there seems to be little agreement about what the "Copenhagen interpretation" is. I don't know a lot about Heisenberg's beliefs, but Bohr's ...
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  • 17.5k
7 votes

Why are wavefunction collapses instantaneous?

A quantum system is always observed to be in a single state, but that doesn't mean that anything has "happened" to the wave function. It's just a point of view. Replace Schrodinger's cat ...
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