Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 156895

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 $\times$, $+$!

0 votes

Why do different ways of computing $\langle p^2 \rangle$ require integration by parts to match?

They aren't different - by utilizing integration by parts, you are explicitly demonstrating that your final and penultimate expressions are equal to one another.
J. Murray's user avatar
  • 71.4k
1 vote

Why are these unbounded operators (essentially) self-adjoint?

On this domain, a great many operators of interest can be shown to be essentially self-adjoint. Consider the restriction of $H=a^\dagger a$ to this domain. …
J. Murray's user avatar
  • 71.4k
2 votes

Given a self-adjoint operator $A$, how does one calculate $d\Gamma(A)$?

I would imagine that the paper in question is using a mild abuse of notation. In general, one has that $$\mathrm d\Gamma(A) \phi_1 \otimes \ldots \otimes \phi_n = \big(A\phi_1\big) \otimes \phi_2 \oti …
J. Murray's user avatar
  • 71.4k
3 votes

Doubts regarding Quantum Mechanics

Measurements are not modeled by the actions of operators on state vectors. [...] So does $|\psi(p)|^2$ tells us about the probability of having the momentum p at time of measurement? … If true, can we generalize it to other operators as well? Yes to both questions. For Example,There is a normalized wavefunction $\psi(x)$, does each x describes the state of particle? …
J. Murray's user avatar
  • 71.4k
4 votes

How to take derivative of density operator?

You just must be careful about the commutation of of operators. … Let $A:t \mapsto A(t)$ be a family of operators on some (for now, finite-dimensional) Hilbert space $\mathscr H$ which are indexed by a continuous variable $t$. …
J. Murray's user avatar
  • 71.4k
3 votes
Accepted

What went wrong in the following calculation of $\langle p'|[x,p]|p'\rangle$?

The wrong step was in assuming that those expressions are well-defined in the first place. The "generalized" bras/kets $|p\rangle$ and $|x\rangle$ are not true members of the Hilbert space, and expres …
J. Murray's user avatar
  • 71.4k
3 votes
Accepted

How to write the density matrix for a state in the basis $\{ |H\rangle,|V\rangle \}$?

Besides combining the $|H\rangle\langle H|$ terms, there's nothing to reduce. If we use matrix notation such that $$a|H\rangle + b|V\rangle \leftrightarrow \pmatrix{a\\b}$$ then generic operator take …
J. Murray's user avatar
  • 71.4k
2 votes

What is the value of the commutator $[L_x, S_y]$?

The operators $L_n$ on $L^2(\mathbb R^3)$ are the familiar differential operators which correspond to the angular momentum associated to spatial wavefunctions, while the operators $S_n$ on $\mathbb C^2 … momentum operators. …
J. Murray's user avatar
  • 71.4k
3 votes
Accepted

How can we show that spin is the generator of rotation?

action of a rotation $R$ on elements of the form $\psi\otimes \phi$ is $$R\big(\psi\otimes \phi\big) = \big(R_A\psi\big) \otimes \big(R_B \phi\big)$$ where $R_A$ and $R_B$ are the corresponding rotation operators … are understood to act on the spatial wavefunction and the $S$ operators are understood to act on the $2s+1$-dimensional spinor. …
J. Murray's user avatar
  • 71.4k
4 votes
Accepted

Physical interpretation of the bra-ket notation

Now, using this bra-ket notation we can compute the inner product of some operator, say $\hat{H}$, so $\langle\psi|\hat{H}|\psi\rangle$ defines the eigenvalue of some hermitian operator $\hat{H}$. T …
J. Murray's user avatar
  • 71.4k
1 vote
Accepted

Is sum of two operators hermitian?

Your operator $A$ is Hermitian if, for any two vectors $\phi_1$ and $\phi_2$, you have that $$\langle \phi_1,A\phi_2\rangle = \langle A\phi_1,\phi_2\rangle$$ In bra-ket notation, this is $$\langle \ph …
J. Murray's user avatar
  • 71.4k
6 votes
Accepted

Are all operators in Quantum Mechanics both Hermitian and Unitary?

Observables in quantum mechanics are represented by Hermitian operators (or rather, self-adjoint operators, though the distinction is more technical than the level of this question), which are not generally … Some operators - such as the ladder operators $a$ and $a^\dagger$ which you will meet when you study the quantum harmonic oscillator - are neither; other operators such as the spatial inversion operator …
J. Murray's user avatar
  • 71.4k
1 vote

Power Series Expansion of Unitary Operators in Weinberg

It's just the multivariable Taylor expansion. For functions of a single variable, $$f(x) = f(0) + x f'(0) + \frac{1}{2} x^2 f''(0) + \ldots$$ For functions of two variables, $$f(x,y) = f(\mathbf 0) …
J. Murray's user avatar
  • 71.4k
2 votes
Accepted

Applying a bra on the density operator

$\langle k|i\rangle$ is just a complex number, and can be moved around freely. In the same way that $3\langle i| = \langle i |3 $, you have that $\langle k|i\rangle\langle = |i\rangle \langle k|i\ran …
J. Murray's user avatar
  • 71.4k
1 vote
Accepted

Complex conjugate and expectation values in QM

A free particle has the wave function $\Psi(\vec{r}) = Ne^{r\gamma}$ where N is a normalisation constant, $\gamma$ is a positive real parameter and $r = \sqrt{x^2+y^2+z^2}$ is the distance from the o …
J. Murray's user avatar
  • 71.4k

1
2 3 4 5
8
15 30 50 per page