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Questions tagged [observables]

A quantum observable is a measurable operator whose corresponding property of the state can be determined by some sequence of physical operations ("observation"), such as submitting the system to various electromagnetic fields and eventually reading a value. In systems governed by classical mechanics, any experimentally observable value can be shown to be given by a real-valued function on the set of all possible system states.

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What if Schrödinger's cat's meowed?

Sorry if this has been asked (every similar question has a title that basically tags Schrödinger's cat) If after the superposition of the cat being dead and alive at one time was created, and the ...
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Expansion of the infinite square well [on hold]

I was studying the expectation value of the energy of a particle in the groud state of the infinite square well after its expansion in terms of width (from $a$ to $2a$), which is: $$\langle H\rangle= ...
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Significance when expectation of a commutator is zero

It is clear to me what it means when the commutator of two operator $[A, B]$ is zero and what it implies. However, is there any significance when the expectation of the commutator $\langle[A, B]\...
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What is 'definite' variable in QM?

I have gone through a few of the questions on the website regarding this particular query, but I have not understood what they meant. When a question says that a particle has definite momentum, are ...
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60 views

Gaussian State Spread [closed]

A measurement device which can be represented by a 1D quantum system (with canonical observables $X$ and $P$) 'is prepared in a Gaussian state with spread $s$' $$\vert \psi \rangle = \frac{1}{(\pi^2s^...
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61 views

Is the period a physical observable in General Relativity?

I am currently seeing the classical tests of GR. To justify the introduction of a test based on the Doppler effect, the professor says that the previous test ( Shapiro and echo-radar test ) is based ...
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Is there any ambiguity when forming operators for classical observables from other operators? [duplicate]

There are some classical quantities, for which we know corresponding operators: e.g. position, momentum. For some others it's straightforward to compose these operators using expressions similar to ...
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587 views

What does vector operator for angular momentum measure?

Consider the vector operator for angular momentum $\hat L=\hat L_x \vec i +\hat L_y \vec j + \hat L_z \vec k$. Does this mean that if we want to measure the angular momentum of a particle in state $\...
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How to define an Operator Product Expansion (OPE) on arbitrary Riemann surface for a CFT?

Whenever we define the OPE of a 2D CFT, we do so (at least in the literature that I have come across) on the complex plane. Similarly, the commutation relations between conformal generators $L_n$ and ...
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1answer
107 views

Simultaneous measurements in Quantum Mechanics

In the notes I am using, it states that if two observables A and B are measured simultaneously, then the measurement of A does not affect the measurement of B, and vice versa. However, why does the ...
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1answer
94 views

What does Ehrenfest's theorem actually mean?

I am told that Ehrenfest's theorem, applied to a physical observable $\hat A$, is: $$\frac{d\langle\hat A\rangle}{dt}= \frac{i}{\bar h}\langle[\hat H,\hat A]\rangle$$ I don't understand how to use ...
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Why does gauge invariance HAVE to correspond to an observable?(Or is it the other way round)

Under the line integral of the geometrical Berry phase, a close-loop integral is gauge invariant as if we were to perform a gauge transformation of the initial state, with the end point of the path in ...
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Why are we only interested in unitary/anti-unitary transformations of the underlying Hilbert space when considering symmetries in quantum mechanics?

Background to question: We briefly looked at 'symmetries' in my quantum mechanics course. I was dissatisfied with the fact that we only considered unitary (touched on antiunitary) operators when ...
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Proof that compatible observables must share a common eigenbasis

Let $A$ and $B$ be two observables, that we suppose bounded and with pure discrete spectrum for simplicity. Let $|{\psi}\rangle$ be a eigenstate of $B$ with eigenvalue $b$. We define $A$ and $B$ as ...
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Does any basis belong to a set of eigenfunctions of some observable?

In the Quantum Mechanics courses in ocw.mit.edu it is said that any wave function can be expanded as a superposition of the eigenfunctions of any observable, i.e the eigenfunctions form a eigenbasis ...
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How come there are Schrödinger Picture operators with explicit time dependence?

In the Schrödinger picture, observables are said to be time independent (see Cohen, for example) operators. However, when deriving the Heisenberg Equation of Motion $$i\hbar\frac{d}{dt}A_H(t)=[A_H(t),...
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Are eigenstates of the position operator continuous?

The way I've understood it is that eigenfunction of an operator are the different states which the actual wavefunction can take when the property/observable corresponding to the given operator is ...
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What does it mean for 2 observables to be compatible?

If I have 2 observable operators $A$ and $B$, if $A$ and $B$ commute: $[A, B] = 0$, then they must necessarily form a complete set of commuting observables (CSCO). Essentially, if 2 observables are ...
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2answers
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Is a quantum gate different from taking a measurement?

I'm reading a book on quantum computing. It is a very non-technical book, and I do not need a very technical explanation. I keep on seeing the words quantum gate pop up, and I'm wondering whether this ...
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Can the complexity of every observable be quantified this way?

In quantum field theory, we have observables localized on all scales. We can construct complicated large-scale observables by multiplying lots of mutually commuting small-scale observables. Some ...
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1answer
85 views

Finding measurements in non-Hermitian operators

I know how the measurement postulate in quantum mechanics works, in regard to hermitian operators, but what if an operator is non-hermitian? Can i apply the following reasoning? If an operator is ...
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1answer
121 views

Heisenberg choice of the observables and different outcomes [duplicate]

Somebody could help me to clarify how its possible that different choices of the observable to measure can lead to different outcomes of the observed system state?
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Are there any ways to exclude uncertainty in the values of any non-commuting operators? [closed]

If two similar systems are created and In the first system the position is measured with accuracy and in the second one the momentum is measured with accuracy can this avoid the uncertainty in the ...
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Does the mean value change?

We know that $x_0$ and $p_0$ are the mean values for the position and momentum of a particle in the normalized state characterized by the function $\psi (x)$, ( that is, $x_0=\langle x \rangle_\psi$ ...
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5answers
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Why is there a physical preference to real numbers?

In quantum mechanics, operators can only be observables if the eigenfunctions they operate on have real eigenvalues. If they are complex, I am told that, surely, some observable of a physical system ...
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Definitions of operators and commutativity in quantum mechanics

If $[\hat A,\hat B] = 0$, where $\hat A$ and $\hat B$ are operators, then the operators commute. This also means that, when applied to a wavefunction, that one can measure observables $A$ and $B$ in ...
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Do all well-measured observables effectively commute?

Do all well-measured observables effectively commute with each other? The rest of this long post clarifies what I mean by that simple-looking question. Consider quantum field theory in Minkowski ...
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4answers
123 views

Is there a concept of velocity at quantum level? [duplicate]

I am confused as according to Heisenberg's uncertainty principle, we cant define the position and momentum of a particle with absolute accuracy simulaneously. Then how can we define velocity at ...
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1answer
48 views

Measuring different observables simultaneously with arbitrary precision

I am trying to understand how different observables can be measured at the same time with arbitrary precision. To check if I understand it I am using an example. Let's say we have as our first ...
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6answers
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If I was Schrödinger's cat, what would I feel? [closed]

What I'm doing Note before reading: I've made two edits for clarification The first starts at: "To clarify based on answers, I think ya'll are missing the meat of the question:" The second starts ...
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1answer
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What happens when the same observable is measured twice in a row?

I have the understanding that the outcome of measuring the same observable twice in a row on the same state is getting the square of the eigenvalue if and only if the state is an eigenstate of the ...
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2answers
438 views

The location of an object is gauge dependent. Therefore, it's not measurable?

The location of an object $x$ depends on how we choose our coordinate system. If we move the zero point, $x$ also changes. However, since we have translational invariance, we can always do such shifts ...
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Is there a quantum analogue of Mean Value Theorem theorem?

Background I was thinking of Mean Value Theorem in the context of classical mechanics I have $2$ points $A$ and $B$ and my particle goes from $A$ to $B$ then I know the velocity of the particle at a ...
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Significance of eigenvalues of an observable Of a wavefunction [closed]

What really is meant by eigenvalue of an observable? Does it mean that everytime we measure a value of an observable the result obtained is equal to the eigenvalue of the observable?
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2answers
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What is a quantum number in a quantum field theory?

In non-relativistic quantum mechanics, quantum numbers are associated with eigenvalues of an operator. For example, $\ell$ is a quantum number associated with the eigenvalue $\ell(\ell+1)\hbar^2$ ...
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Under what circumstances are general relativistic coordinate transformations physically meaningful?

Although the field equations of GR are covariant under arbitrary coordinate transformations, such as the transformation given by Dirac (in Princeton Landmarks pp 34) that eliminate the singularities ...
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1answer
38 views

Relationship between the Galilei Group and the Phase Space

This question is kind of a follow up question to my last question on the need for canonical commutation relations and conjugate observables. A comment from Valter Moretti suggested that, given a ...
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2answers
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I have a question about momentum and energy of the infinite square well in quantum mechanics

In Griffiths quantum mechanics, There is a problem that "Find the momentum-space wave function $\varphi(p,t)$ for the $n$th stationary state of the infinite square well." The $n$th stationary ...
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2answers
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Observer's effect and the Heisenberg choice

Quantum Mechanics postulates that the act of observation affects the behavior of the observed object.The most common example of this feature is the fact the unobserved object(e.g. a photon) behave ...
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1answer
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Average value of non-projective observables

I am quite confused about how to measure observables (like Pauli spins). For example, in the exercise 2.66 of Nielsen and Chuang's textbook: Show that the average value of the observable $X_1Z_2$ ...
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1answer
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Confusion about why deducing pointer observable from the structure of the Hamiltonians is not practical

I am trying to learn Zurek's theory of decoherence. Right now I am reading Decoherence, einselection and the existential interpretation (the rough guide) which seems like an easier read than his big ...
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7answers
201 views

What do positions in Schrodinger Equation mean (Remember: the particle never has definite position)?

In position or configuration representation, the Hamiltonian operator, and thus the Schrodinger Equation, is expressed in terms of positions. But the particle never has definite position, what do ...
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4answers
207 views

Is there a way to measure this observable in QM?

Let a quantum system be described by Hilbert space $\mathscr{H}$ and let $|\psi\rangle$ be an arbitrary state. Define the operator $$P=|\psi\rangle\langle \psi|$$ This is hermitian. It has two ...
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1answer
108 views

What is the physical meaning of expectation value of the Hamiltonian operator?

I've been studying David Griffiths' Introduction to Quantum Mechanics and int that, it was explained that the expectation value of position $x$ is the average of the positions of $N$ identically ...
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A moment of cohomology$.$

As is well-known (cf. Ref.1), the momentum operator is defined up to a time-independent closed form. More precisely, the physically inequivalent momentum operators are classified by the de Rham ...
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1answer
178 views

No sense in the expression $\hat{x}| 1\rangle=\sqrt{\frac{2}{a}}\int_{-\frac{a}{2}}^{\frac{a}{2}}x\cos\left(\frac{\pi}{a}x\right)dx=0$

I am considering a particle of mass m in a symmetric infinite square well of width a in the fundamental state. $$V(x)= \begin{cases} 0 & \mbox{$|x|<\frac{a}{2}$} \\ \infty & \mbox{...
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Why do we say spin/angular momentum is observable even though its components can't be determined simultaneously?

Why do we say spin or angular momentum of a particle is observable even though all of its components can't be determined simultaneously? For example, we can measure the $\hat{L_x}$ of a particle's ...
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Quantum Observation

Bear with me if I present a lack of knowledge - QM is not my field. There's a common notion in QM that until a particle is observed (measured), its properties are not definite, but rather are spread ...
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Identify a $|Ψ(0)⟩$ with $A|Ψ(0)⟩≠a|Ψ(0)⟩$ $\forall$ $A$ & $|Ψ(0)⟩=\sum a_j\lvert\chi_j\rangle$ for some $A$ and its eigenstates$\lvert\chi_j\rangle$

Is it possible to put a quantum system in a state at time $t=0$, which is not the eigenstate of any observable, but at the same time can be linearly expanded using the eigenstates of some observable? ...
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Doubt about fifth postulate of QM degenerate case

I think I didn't not really understand a comment found on "Quantum Mechanics" by Claude Cohen-Tannoudji. Talking about the fifth postulate of quantum mechanics in the case in which $a_n$ is a ...