# Tag Info

37

In short terms, kets are vectors on your Hilbert space, while bras are linear functionals of the kets to the complex plane $$\left|\psi\right>\in \mathcal{H}$$ \begin{split} \left<\phi\right|:\mathcal{H} &\to \mathbb{C}\\ \left|\psi\right> &\mapsto \left<\phi\middle|\psi\right> \end{split} Due to the Riesz-Frechet theorem, a ...

28

I would like to extend Alex' answer, as well as answer your question in the comments: "Is the "bra" basically something like the dot product?" If you have a vector space $V$ over a field $F$, which is, for now, finite dimensional, you can create another vector space, $V^*$, called the dual space of $V$, which consists of linear functionals defined on $V$. ...

28

First, the $bra$c$ket$ notation is simply a convenience invented to greatly simplify, and abstractify the mathematical manipulations being done in quantum mechanics. It is easiest to begin explaining the abstract vector we call the "ket". The ket-vector $|\psi\rangle$ is an abstract vector, it has a certain "size" or "dimension", but without specifying ...

11

Here's a pretty direct hint. If there were only rationals, the function could be defined as the denominator in lowest terms, because everything nearby enough would have a larger denominator. If we want to also define it on the irrationals this wouldn't work. However, we can compress the positive integers into $[0,1)$ in an order preserving way, then we can ...

11

I think that the function that sends any irrational number on $1$ and sends a rational $\frac{p}q$ (where the fraction is irreducible) on $1-\frac1q$ does the job. Given any rational number $\frac{p}q$, you can always find a neighborhood so that any rational number has a denominator bigger than $q$

10

I would not worry too much at first about distinguishing between "bra" and "ket". What is important is that you have an inner product $\langle x | y \rangle$ which is linear in the second coordinate and conjugate linear in the first, as opposed to the Mathematician's inner product $(x,y)$ which is linear the first coordinate and conjugate linear in the ...

7

What helped me understand it was the notion of bra and ket as vectors in Hilbert space. ket $|f\rangle$ denotes a "usual" vector, bra $\langle x|$ a "transposed" one which can be used for projection. Thus $$\langle x | f \rangle = f(x)$$ is simply the projection of $f$ into its (coordinate) representation of the $x$. This becomes more clear when you see ...

6

The proof has nothing to do with the Schwartz space per se; nor with $i$ or $t$, or that $P$ and $Q$ are symmetric. If $P,Q$ are operators on a Hilbert space $H$ with domain $D$ and such that $PD\subset D$, $QD\subset D$, and $QP-PQ=\mathbb I$, then at least one of $P$ and $Q$ is unbounded. This applies to the case in the question because if we have ...

5

I’d attack it much more directly. HINT: Suppose that $a=\langle a_n:n\in\Bbb Z^+\rangle\notin\ell_\infty$. Then $a$ has a subsequence $\langle a_{n_k}:k\in\Bbb Z^+\rangle$ such that $|a_{n_k}|\ge k$ for each $k\in\Bbb Z^+$. For each $k\in\Bbb Z^+$ let $$x_{n_k}=\frac1{ka_{n_k}}\;,$$ and let all the other terms of $x$ be $0$. Show that $x\in\ell_1$, ...

5

I'll assume $S$ not empty. Since the function $f\colon S\to\mathbb{R}$, $f(p)=d(p_0,p)$ is continuous, when $S$ is compact its image is compact, hence closed and bounded; therefore the image of $f$ contains its minimum. If $S$ is only assumed to be closed and bounded, but not compact, the statement is not generally true. Consider $X=\{0\}\cup (1,2]$, with ...

5

Take $T_k x = k^2 x_1 -k x_2$. Then for any $x \neq 0$ we see that $|T_kx| \to \infty$. However, $T_k ({1 \over \sqrt{1 + k^2}}(1,k)) = 0$ for all $k$.

5

It fails for the sequence $1,0,0,\dots$

5

It's not quite as simple as that. The union $$N = \bigcup_{j,k,m \in \Bbb{N}} N_{j,k,m}$$ is a countable union, so properties of measures hold. If we take the most straightforward implementation of your replacement scheme, we need $$N = \bigcup_{\substack{j,k\in \Bbb{N} \\ 0 < \varepsilon < \varepsilon_0}} N_{j,k,\varepsilon}$$ which is no ...

5

If the $n$ real-valued continuous functions $f_1, \ldots, f_n$ separate points of $K$, then $(f_1, \ldots, f_n)$ is a homeomorphism from $K$ to a compact subset of $\mathbb R^n$. But not every compact metric space is homeomorphic to a compact subset of $\mathbb R^n$. For example, let $K$ be the Hilbert cube. For each $k$, $K$ has a subset $S_k$ ...

4

Well, functional analysis provides very natural set up for dealing with this question. We will prove the same for $$A_{c(x)}:= \left \{p(x) e^{-c(x)} \right\}$$ Where $c(x)$ is a fixed even degree polynomial with positive leading coefficient and of degree $\ge 2$ and $p(x)$ is any polynomial. Note that we need to prove that $A_{c(x)}$ is dense in $C_0 ... 4 Notice that for each vector$x$, one has $$\|T^* T^2(x)\|^2 = \langle T^* T^2(x),T^* T^2(x) \rangle = \langle TT^*T^2(x), T^2(x)\rangle = \langle T^*T^3(x),T^2(x) \rangle = \langle T^3(x),T^3(x) \rangle = \|T^3(x)\|^2.$$ Thus $$\|T^3\| = \sup_{\|x\|=1} \|T^3(x)\| = \sup_{\|x\|=1}\|T^*T^2(x)\| = \|T^*T^2\|.$$ 4 We know that for$\|A\| < 1$,$(I-A)^{-1}$is well-defined (prove this yourself if you have not done so yet) so we can talk about the inverse. Thus: $$I = (I-A)(I-A)^{-1}.$$ Here is a hint: $$1 = \|I\| = \|(I-A)(I-A)^{-1}\|.$$ Try doing some basic norm manipulations to this. You need to increase the norm, not decrease it since you want a lower ... 4 Start with an orthonormal set$\{ f_n \}_{n=1}^{\infty}$in$L^2[1/2,1]$, and extend the functions to be$0$on$[0,1/2]$in order to obtain an orthonormal set$\{ \tilde{f}_n \}_{n=1}^{\infty}$in$L^2[0,1]$. Then, for$n\ne m, \begin{align} \|T\tilde{f}_n-T\tilde{f_m}\|^2 & =\|x\tilde{f}_{n}-x\tilde{f}_m\|^2 \\ & ... 4 IfA$is symmetric (Hermitian), then that's true (assuming you're talking about the induced Euclidean norm). However, in general,$\|A\|$can be very large. For example, take $$A=\pmatrix{1&t\\0&-1}$$ with$t$as large as you want to make it. Note that for any operator on a Hilbert space, we have$\|A^*A\| = \|A\|^2$(you should have this as a ... 4 Suppose you are trying to solve$Af=g$where$A$is a linear operator. As always, there are two big issues: existence and uniqueness. Start with a Hilbert space for the sake of discussion. Existence: A solution of$Af=g$exists iff$g$is in the range of$A$. If$A$happens to have a closed range, then the issue is whether or not$g$is in the range is ... 4 When dealing with conjectures about unbounded operators, it's always good to test conjectures with a differential operator. John von Neumann defined closed unbounded operators to study differential operators. Differential operators are still the best examples. For example, let$X=C[0,1]$, and let$A=\frac{d}{dx}$on the domain$\mathcal{D}(A)$of ... 4 Use paralleogram law for$\frac{x}{2}$and$\frac{y}{2}$to obtain$||\frac{x+y}{2}||^2 + ||\frac{x-y}{2}||^2 = \frac{1}{2}||x||^2 + \frac{1}{2}||y||^2$and so you get$1$in the right hand side. Since the LHS is a sum of two non-negative terms, you get the desired inequality since$x\neq y$. 4 The accepted answer is incorrect. To show that$S$is closed, you must show that for any sequence of points$(x_n)$in$S$which converges to a limit$x\in \mathbb{R}$, the limit$x$is also in$S$. You can prove this for your$S$by contradiction: suppose$x_n\to x$but$x\not\in S$. Now use the definition of limit (with$\epsilon=-x$) to show that ... 4 The first thing I thought of was a "soft proof": Suppose$f \to f'$maps$H^\infty$to$H^\infty.$By the closed graph theorem, this linear map is continuous. Thus there exists$C$such that$\|f'\|_\infty \le C\|f\|_\infty$for all$f\in H^\infty.$The functions$z^n$show this fails. But it's easier to just note$f(z) = \sum_{n=1}^\infty z^n/n^2$is a ... 4 The only reason the question seems silly is that you include the answer! A locally convex TVS is one that has a basis at the origin consisting of balanced absorbing convex sets. The reason for the emphasis on "convex" is that that's what distinguishes locally convex TVSs from other TVSs: every TVS has a local base consisting of balanced absorbing sets. ... 4 All you need to do is a little algebra to substitute$f_1 + f_2$into the new definition. To apply Minkowski inequality, let's denote the traditional$L_2$norm of$f$:$\left(\int_0^1 |f|^2 dx\right)^{1/2}$by$\|f\|_0. We have: \begin{align} & \|f_1 + f_2\|^2 \\ = & \int_0^1 |f_1 + f_2|^2 dx + \int_0^1 |f_1' + f_2'|^2 dx \\ = & \|f_1 + ... 4 Suppose you have a non-trivial solution of $$T^*Tf = \int_{t}^{1}\int_{0}^{s}f(y)dy ds = \lambda f(t)$$ Then\lambda \ne 0$because the above would give$f=0$after differentiating a couple of times. For$\lambda \ne 0$, any solution of the above must satisfy $$\lambda f'' = -f \\ f(1)=0,\;\; f'(0)=0.$$ Any ... 3 Given$f$and$\epsilon$, choose a polynomial$p$with$\Vert f-p\Vert_{\infty,X}<\epsilon$(where$\Vert\cdot\Vert_{\infty,X}$is the supremum norm oin$X$). Now see the corresponding polynomial function in$\mathcal{A}$,$p:\mathcal{A}\to\mathcal{A}$. (Remember: the functional calculus respects this notation, i.e.,$p(a)$, in the functional calculus, is ... 3 This is studied in potential theory: the function$u$is the Newtonian potential of$f$, $$u(x)=\int_{\mathbb{R}^n} K(x-y)f(y)\,dy$$ where$K(x)=c_n|x|^{2-n}$for$n\ne 2$and$K(x)=c_2\log|x|$for$n=2$. In dimensions$n\ge 3$the kernel$K$decays at infinity, so$u(x)\to 0$as$|x|\to\infty$in this case, provided$f$is reasonable (integrable and ... 3 Define$\phi:(B+I)/I\to B/B\cap I$by $$\phi(b+j+I)=b+B\cap I,\ \ \ \ \ b\in B,\ j\in I.$$ Of course we need to check that this is well-defined. If$b_1+j_1=b_2+j_2$, then $$b_1-b_2=j_2-j_1\in B\cap I,$$ so$b_1+B\cap I=b_2+B\cap I$. The map is obviously linear, multiplicative,$*$-preserving, and onto. As for injectivity, if$b_1+B\cap I=b_2+B\cap I\$, ...

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