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Results tagged with operators
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user 326183
In physics, an operator is almost always either a square matrix or a linear mapping between two function spaces (defined on, say, $\mathbb R^n$). Operators serve as observables and as time evolution operators in Quantum Mechanics. This tag will most often find valid use in quantum mechanics; don't use this tag just because your equations contain "everyday operations" like $\times$, $+$!
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How can an operator be proportional to a scalar?
I am an undergraduate physics student reading through some parts of Griffiths's Quantum.
I recently saw that $k$ is proportional to momentum $p$ via the De Broglie relation. But, to my understanding
$ …
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2
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Are all operators in Quantum Mechanics both Hermitian and Unitary?
Operators in general can be Hermitian, unitary, both, or neither. … In particular, observables are usually described using Hermitian $\text{operators}.$ And, transformations are described using unitary operators. …
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Non-finite expectation values in quantum mechanics [duplicate]
In textbook quantum mechanics, one deals with expectation values of the form
$$\langle O \rangle = \text{tr}(\rho O)$$
where $\rho$ is assumed to be trace-class (in particular, $\text{tr}\rho = 1$). H …
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Question regarding Hermitian property of Operators
It looks like you are mixing notations. Let us clarify a few things.
Let $\mathcal{H}$ be our Hilbert space. Let $\hat{A}$ be an operator over our Hilbert space, i.e., $\hat{A}: \mathcal{H} \rightarro …
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Physical justification for $SU(2)$ being the non-relativistic spin group
To preface, I have little background in representation and Lie theory.
My understanding is as such: Given any finite dimensional Hilbert space $\mathcal{H}$ and representation $\rho: SU(2) \rightarrow …
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3
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198
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What is the mathematically precise definition of raising and lowering operators?
Is there an "axiomatic" definition of raising and lowering operators of a given operator (e.g. spin in the $z$-direction or quantum harmonic oscillator Hamiltonian)? … I am particularly interested in understanding when raising and lowering operators can be defined for a given operator. …
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What, if anything, is the generator of point inversion?
Nothing is the generator of point inversions. Hereafter, I will explain my reasoning.
Consider a Lie group $G$ and its Lie algebra $\mathfrak{g}$. An element $A \in G$ is generated by its Lie algebra …
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What are the distinct mathematical formalisms of quantum mechanics?
Observables are linear operators $\hat{O}: \mathcal{H} \to \mathcal{H}$ that live in a particular algebra endowed with two multiplication operations (multiplication and a Lie bracket). …
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What is the purpose of finding the eigenvalues and eigenvectors of a hamiltonian?
If you're dealing with the time-independent Schrödinger equation (TISE), then you get the following eigenvalue problem:
$$H\Psi = E\Psi$$
where we define $E$ to be the energy of the state.
Thus, its e …
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2
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Unitary Time Evolution Operator
.$$
This seems to imply to treat the exponentials (the unitary time-evolution operators) as constants. …
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Accepted
What does the finding the eigenvalue of a wavefunction physically mean?
To somewhat see this, we must talk about observables and their corresponding operators. … These operators are usually denoted as operators with a little hat. For example, there is a position operator $\hat{x}$, a momentum operator $\hat{p}$, and an energy operator $\hat{H}$. …
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How can I derive the fact that there are no "non-integral" raising and lowering operators fo...
One crucial piece of this argument is the introduction of raising and lowering operators:
$$S_{\pm} = S_x\pm iS_y,$$
which raises/lowers the $S_z$ eigenvalue of a state by $\hbar$. … What I do not understand is: How can I prove that no other operators exists which raise or lower the $S_z$ eigenvalue of a state by a non-integral unit of $\hbar$? …
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Accepted
Quantum mechanics: Can you simplify $\langle x\rangle\langle p\rangle$ further?
Firstly,
$$|\psi \rangle \langle \psi |\neq 1 \tag{1}$$
in general. The operator (1) will project a state onto the $|\psi\rangle$ state.
Another way that you can see that your manipulation is wrong is …
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Accepted
Commuting observables and an eigenstate
If $A$ and $B$ have nondegenerate spectrums, $[A, B] = 0$ implies that they share a unique set of eigenvectors which form a basis for the Hilbert space said operators are defined over. … a_1, b_1 \rangle$ which makes it clear that the ket is an eigenstate of both operators with the corresponding eigenvalues. …
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The Quantum theory of non-relativistic angular momentum in representation theoretic terms
I am trying to understand the Quantum theory of non-relativistic angular momentum in terms of representation theory with full precision. In particular, I would like to deduce that angular momentum is …
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