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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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Is there a difference between a Hermitian operator and an observable?

My poorly written lecture notes say that any Hermitian operator does have a complete set of orthogonal eigenstates with real corresponding eigenvalues and is therefore an observable. In the article ...
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Why is an operator the quantum mechanical analogue of an observable?

I used to think because that, if objects are treated as waves, then using operators is the necessary thing to do in order to "retrieve" the observable from a given wavefunction. For example, in ...
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Observing the conserved canonical momenta

Suppose I have a Lagrangian $\mathcal{L}[\phi]$ with $\phi$ a cyclic variable, which means that the Lagrangian is symmetric under shift of $\phi\rightarrow\phi+c\quad$. The equation of motion will be ...
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Quantum mechanics on operator [closed]

If any operator is commute with Hamilton then they are labelled such a way that the energy eigenstate are equal and we also know it is a constant of motion. I don't related constant of motion with ...
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Measurement formalisms - POVM formalism vs Hermitian observables

I am thinking in following way of thinking about measurements in quantum mechanics. Please correct any false statements I may be making below. We start with POVMs. Let our POVM be a set of positive ...
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Meaning of time derivative of an operator

Today when my professor was deriving this equation: $$\frac{\mathrm d\langle A\rangle}{\mathrm dt}=\frac{i}{\hbar}\langle\left[H,\,A\right]\rangle+\left\langle\frac{\partial A}{\partial t}\right\...
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118 views

“Commuting observables share common eigenstates”

I am struggling to find a precise definition of this line from my quantum mechanics textbook: If $[A,B] = 0$, then the operators commute, and "commuting operators share common eigenstates". This ...
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What are the orthogonal eigenstates of the field operator?

In Peskin & Schroeder section 9.2, they derive the two-point function in the path integral formalism: $$\langle \Omega | \mathcal{T} \left\{ \hat{\phi}(x_1)\hat{\phi}(x_2)\right\} | \Omega \...
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Are $\hat x$ and $\hat p$ assumed to be time-independent operators?

In the book Quantum Mechanics by Cohen-Tannoudji, at $G_{III}$, it is given that and then in the comment section, it is also given that so I'm pretty confused in here, because in one side, they say ...
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For any two unitarily equivalent observables, can both be measured by the same experimental apparatus?

If we have an observable $A$, and a unitary operator $\hat U$, one can easily show that both $\hat A$ and $\hat U \hat A \hat U^{\dagger}$ have the same spectrum - in fact, they are called unitarily ...
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is there a way to experimentally determine the mean of $\hat A$, namely $\langle \hat A \rangle $?

Let $A$ be an observable, then, is there a way to experimentally determine the mean of $\hat A$, namely $\langle \hat A \rangle $? I mean, for example, consider the position operator; it is ...
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1answer
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Is the identity operator correspond to the observable which states the existence of system?

As it is claimed in this question, the identity operator is an hermitian operator, but not an observable. However, if I were to build a device, which only measures the existence of an electron in a ...
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Is there any intuitive reason behind why should the eigenfunctions of observables form a basis for our Hilbert space?

Is there any intuitive reason behind why should the eigenfunctions of observables form a basis for our Hilbert space ? For example, in the case of Stern-Gerlach experiment, sending the beam that has ...
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Is is possible to have a pair commuting observables only in a single direction?

In quantum mechanics, for two observables to be compatible, successive measurements of the observables, say $A$ and $B$, should yield the same result as earlier, i.e if we do the measurements with the ...
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What is the variance of |S| in Bell's inequality (CHSH inequality)

Sorry that this isn't a quick question but I didn't know how to make it shorter. I am struggling with this for quite a long time and I would appreciate every help that I can get. I could not find a ...
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How to write an operator in matrix form?

Say I have the following operator: $$\hat { L } =\hbar { \sum_{ \sigma ,l,p } { l } \int_{ 0 }^{ \infty }\!{ \mathrm{d}{ k }_{ 0 }\,\hat { { { a }}}_{ \sigma ,l,p }^{ \dagger } } } \left({ k }_{...
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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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142 views

Expansion of the infinite square well [closed]

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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1answer
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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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1answer
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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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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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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
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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
158 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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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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339 views

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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1answer
147 views

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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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
139 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
132 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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1answer
127 views

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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2answers
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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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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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1answer
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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
200 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
71 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
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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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1answer
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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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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 $\ell(\ell+1)\hbar^2$ ...