# Tagged Questions

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### Superposition and simultaneous observation

Trying to understand superposition. Ok, so double slit experiment. The multiple paths the particle simultaneously travels interfere with each other but as it is absorbed, it chooses one "actual" ...
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### Are the authors saying that the observer effect plays no role in Bohr's thought experiment of the Heisenberg uncertainty principle?

Here is an excerpt from Eisberg & Resnick's Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles. Here is introducing Bohr's though experiment to establish a physical origin for the ...
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### Why do we use Hermitian operators in QM?

Position, momentum, energy and other observables yield real-valued measurements. The Hilbert-space formalism accounts for this physical fact by associating observables with Hermitian ('self-adjoint') ...
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### Is commutation relation an equivalence relation?

I'm now learning quantum mechanics with Liboff. In the book it deals with "a compete set of mutually compatible observables" in order to make a state maximally informative. How can one find such set? ...
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### How to recognize a Complete Set of Commuting Operators (CSCO)

A question about 'completeness'. These two operators are commuting, but I want to know more about their completeness. How do you know if {H}, {B}, {H,B} and/or {$H^2$,B} are forming (a) Complete ...
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### Taking Measurements of Quantities in QM

I have a quick question relating to Annihilation and Creation operators, and in taking observables in general. Let's say, for instance, that I prepare a particle so that I consider the projection of ...
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### Trace of an observable [closed]

If $X$ and $Y$ are two observables and $\rho$ is a density operator, is it true that for every complex number $z$ the quantity $$\mathrm{tr}[\rho (X+zY)^*(X+zY)]$$ is non-negative?
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### If a quantum state is pure why are its observables still probabilistic?

As I understand it, a pure quantum state is one that can be represented as a ket $\lvert\psi\rangle$ in a Hilbert space, and it contains all the information about the state of the system. As such, we ...
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### Sequential Stern-Gerlach devices - realizable experiment or teaching aid?

At least one textbook [1] uses sequential Stern-Gerlach devices to introduce to students that the components of angular momentum are incompatible observables. Viz., the $z$-up beam from a SG device ...
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### What is $\langle \phi | H | \psi \rangle$ in QM?

I know that $\langle \phi | \psi \rangle$ is the probability of going from the $\psi$-state to the $\phi$-state, and that $\langle \phi | H | \phi \rangle$ is the expectation value of the energy for ...
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### Physical quantities have definite values?

I don't really know if this question has an anwser but I thought it was worth to try asking. My point here is the following: in Quantum Mechanics, to describe the states of a system we use a Hilbert ...
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### Is there any additional complexity in the physical laws that seems unnecessary for us to exist? [closed]

I am wondering if the universe is as simple as possible, at least given the constraint that humans exist on Earth. This is my second attempt at this question, which was marked as too broad, since I ...
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### Proof of Canonical Commutation Relation (CCR)

I am not sure how $QP-PQ =i\hbar$ where $P$ represent momentum and $Q$ represent position. $Q$ and $P$ are matrices. The question would be, how can $Q$ and $P$ be formulated as a matrix? Also, what is ...
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### Why is only one quantity of angular momentum i.e. $L_z$ quantized & not $L_x$ & $L_y$?

This is quoted from Arthur Beiser's Concepts of Modern Physics: Why is only one quantity of $\mathbf{L}$ quantized? The answer is related to the fact that $\mathbf{L}$ can never point in any ...
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### Have Witten-type TQFT's nonconservation of energy and momentum in interactions?

Witten-type topological quantum field theories are based on cohomology theories. Every observable must lie in a cohomology class. May be $G$ a geometric field. Then every observable expectation value ...
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### Why don't non-Hermitian operators with all real-eigenvalues correspond to observables? [duplicate]

Suppose you could construct an operator that was non-Hermitian but had all real eigenvalues or could at least be restricted in a way to create only real eigenvalues, why would this operator not ...