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Results tagged with operators
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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 $\times$, $+$!
3
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Projectors and Hermitian Operators
Yes. For any vector $\psi$ in the Hilbert space, the projector $P_\psi$, which eats a vector $\phi$ and spits out $\langle\psi,\phi\rangle \psi$, is bounded (with operator norm $\Vert P_\psi \Vert_{o …
- 71.3k
8
votes
Accepted
Is $\hat u \hat p$ also a projector if $\hat p$ is and if $\hat u$ is unitary?
No, it is not true. Projectors are, by definition, idempotent; however, in general $(\hat u\hat p)(\hat u \hat p) \neq \hat u \hat p$. As a trivial example, let $\hat u = -\mathbb I$, which maps $\p …
- 71.3k
1
vote
Accepted
How to show that $\hat{A}-\langle a\rangle$ is Hermitian?
It is also easy to show that if two operators $A$ and $B$ are both Hermitian, then so is their sum/difference. …
- 71.3k
2
votes
Accepted
Are the creation operators on the fermionic Fock space bounded linear operators (do they hav...
Yes, it is bounded with norm $\Vert f \Vert$.
To see this more clearly, note that we are always free to choose an orthonormal basis $\{e_i\}$ of $H$ such that $f = \alpha e_1$ for some $\alpha\in\mat …
- 71.3k
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
$$\mathrm d\Gamma(A) \phi_1 \otimes \ldots \otimes \phi_n = \big(A\phi_1\big) \otimes \phi_2 \oti …
- 71.3k
6
votes
Accepted
What do smeared field operators act on?
The same is true for quantum field operators, where $\hat\phi(x)$ is a singular object for the same reason that $|x\rangle$ is. …
- 71.3k
3
votes
Existence of Hermitian adjoint
This is a subtle issue, and it's certainly not true that any operator which can be written as a "function" of the position and momentum operators is defined on a dense domain. …
- 71.3k
3
votes
Eigenvalue of Orbital Operator
$L^2$ and $L_z$ commute, so it's possible to find a simultaneous eigenbasis of both operators. …
- 71.3k
2
votes
Accepted
Completeness in tensor product basis
Let's say I have an operator A that acts only on the $|\alpha\rangle$ space.
If the Hilbert space under consideration is the tensor product space $\mathcal H = \mathcal H_\alpha \otimes \mathcal H_\ …
- 71.3k
2
votes
Accepted
Question on Dirac's "Principles of Quantum Mechanics"
To be clear, the statement is the following:
Let $|\psi_b\rangle$ be an eigenket of the operator $\zeta$ with eigenvalue $b$ - i.e. $\zeta |\psi_b\rangle = b |\psi_b\rangle$. Define an operator $ …
- 71.3k
0
votes
Demonstration of a property of the adjoint operator in Quantum Mechanics
You might consider the fact that
$$ \langle \alpha | A \ \beta\rangle = \langle A^\dagger \alpha |\beta\rangle = \overline{\langle\beta|A^\dagger \alpha\rangle}$$
Where the line denotes complex conj …
- 71.3k
3
votes
Accepted
Why does $a^\dagger a a^\dagger \left|n\right\rangle = a^\dagger \left|n\right\rangle + a^\d...
Simply note that $[a,a^\dagger]=1 \implies aa^\dagger = 1+a^\dagger a$.
- 71.3k
8
votes
Accepted
How to prove the det of the dot product of a vector and a Pauli vector is minus the vector i...
A cute proof that involves your suggested relation goes as follows. First, note that by definition, $\mathbf a \cdot \boldsymbol \sigma$ is a $2\times 2$ traceless Hermitian matrix. As such, its eig …
- 71.3k
2
votes
Accepted
Energy basis to the X basis
First, we need to lay out some preliminaries in the language of Shankar's book. If you want, you can skip to the bottom where I apply the following to your question.
When you express a state $|\ps …
- 71.3k
5
votes
Accepted
Representation of identity operator
The sum of these two operators is
$$|+\rangle\langle+| + |-\rangle\langle-| = \pmatrix{1&0\\0&0}+\pmatrix{0&0\\0&1} = \pmatrix{1&0\\0&1}=\mathbb I$$
where $\mathbb I$ is the identity operator. …
- 71.3k