All Questions
66
questions
1
vote
0
answers
94
views
Using density argument of Schwartz functions in $L^2$ to show an operator is symmetric
The position operator $X$ is defined as multiplication by $x$, i.e.
$$(X\psi)(x):=x\psi(x).$$
We take the domain $D(X)\subset L^2(\mathbb R)$ for $X$, which we can take
$$D(X):=\{\psi \in L^2(\mathbb ...
- 5,601
2
votes
4
answers
121
views
The value of $\langle p^2 \rangle$ for $\psi(x)=e^{-|x|}$, where $\hat{p}:=i\frac{d}{dx}$ is the momentum operator.
Consider the Hilbert space $H$ of square-integrable complex functions on the real line equipped with the inner product $$\langle \phi,\psi\rangle:=\int \phi^*\psi d\mu,$$ where $\mu$ is the Lebesgue ...
2
votes
1
answer
202
views
Spectrum of a sum of self-adjoint operators
This is a "sequel" to that question where I explain why I need the spectrum of an operator given as the sum of a convolution and a function multiplication. Here, I am considering the ...
- 1,139
2
votes
2
answers
266
views
Compactness of subset of trace-class operators on a Hilbert space
Consider an infinite-dimensional, complex and separable Hilbert space $H$ and let $\mathcal I(H)$ denote the space of trace-class operators.
The set of density operators is defined by $$\mathcal S(H):...
- 428
6
votes
0
answers
311
views
Limit of a particular trace norm.
I have the following problem.
Let $\mathbf{\hat{\rho}}(t)$ and $\mathbf{\hat{\sigma}}(t)$ be two trace class positive operators acting on a Hilbert space of infinite dimension for all $t > 0$. More ...
- 71
1
vote
0
answers
43
views
If $A\geq 0$ and $0\leq B\leq C$, then $0\leq \sqrt{B}A\sqrt{B}\leq \sqrt{C}A\sqrt{C}$ [duplicate]
Let $H$ be a complex Hilbert space with inner product $\langle\cdot,\cdot\rangle$. We say $A\geq 0$ if $\langle Ax,x\rangle\geq0$ for all $x\in H$. If $A\geq 0$ and $0\leq B\leq C$, then does $0\leq \...
- 461
2
votes
1
answer
77
views
If $0\leq A\leq B$ then the corresponding densities satisfy $0\leq\rho_A\leq \rho_B$
Let $A,B$ on $L^2(\mathbb R^d,\mathbb C^d)$ such that $0\leq A\leq B$ AND $A,B$ are trace-class. Then their densities are given by $$\rho_A(x):=\sum_jA\varphi_j(x)\overline{\varphi_j(x)},\qquad \rho_A(...
- 461
0
votes
1
answer
70
views
If $0\leq A\leq B$ and $C\geq0$ then does $0\leq AC\leq BC$?
Let $H$ be a complex Hilbert space with inner product $\langle\cdot,\cdot\rangle$. We say $A\geq 0$ if $\langle Ax,x\rangle\geq0$ for all $x\in H$. If $0\leq A\leq B$ and $C\geq0$ then does $0\leq AC\...
- 461
1
vote
0
answers
34
views
Meaning of "smooth finite-rank operators $Q$ in $L^2(\mathbb R^d)$"
What does it mean for a finite-rank operator to be "smooth"? How can one finite-rank operator be smoother than another? For me, a finite-rank operator on $L^2(\mathbb R^d)$ has an expression
...
- 461
1
vote
1
answer
84
views
Operator theory: nonnegative operator $Q$ less than an orthogonal projection $P$ satisfies $PQ=QP$
Suppose that we have a Hilbert space $\mathcal{H}$, an orthogonal projection $P$ (i.e. $P=|\Psi\rangle\langle\Psi|$ for some $\Psi\in\mathcal{H}$ with $\|\Psi\|=1$) and another non-negative bounded ...
- 85
0
votes
1
answer
40
views
If $A$ is trace-class on $L^2(\mathbb R^2)$ then $\mathbb 1(-\Delta\leq 1)A\mathbb 1(-\Delta\leq 1)$ is finite-rank. [closed]
I am trying to prove that if $A$ is trace-class on $L^2(\mathbb R^2)$ then $\mathbb 1(-\Delta\leq 1)A\mathbb 1(-\Delta\leq 1)$ is a finite-rank operator, where $\mathbb 1(-\Delta\leq 1)$ is defined on ...
- 461
1
vote
1
answer
65
views
An inequality for Schatten norms with a compact self-adjoint operator
Let $Q$ be a compact self-adjoint operator on $L^2(\mathbb R,\mathbb C)$. Notice that $(1-\Delta)^{s}$ is a positive (and hence self-adjoint) operator on $H$ for any $s>0$.
We denote by $\mathfrak ...
- 461
4
votes
1
answer
145
views
Definition of ground state?
In quantum mechanics, a ground state is an eigenstate of the hamiltonian with the minimal eigenvalue and its existence is guaranteed by appropriate theorems.
At least that's how it's defined in ...
- 305
0
votes
0
answers
63
views
What is the most general Self-Adjoint operator acting on $L^{2}(\mathbb{R})$?
What is the most general Self-Adjoint operator acting on $L^{2}\mathbb{R}$?
My hypothesis is that the aswer to my question will be that the most general Self-Adjoint operator $A$ acting on $L^{2}(\...
- 71
1
vote
0
answers
105
views
On the structure of the set of Self-Adjoint operators acting on $L^{2}(\mathbb{R})$.
Let us consider the $Hilbert$ Space $L^{2}(\mathbb{R})$ and let $SA(L^{2}(\mathbb{R})$ be the space of all self adjoint operators acting on $L^{2}$. I have worked with operators such as $X$, $P$, $X^{...
- 71