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21

This question gets to the heart of what makes quantum mechanical amplitudes different from classical probabilities. It is true that if you make measurements in the basis of states $\{\lvert 00 \rangle ,\lvert 11 \rangle\}$ then the two states have the same measurement statistics, and so cannot be distinguished. The interesting thing is that it is possible to ...


16

I understand the definition of a Hilbert space. But I do not understand why non-commutativity compels us to use Hilbert spaces. It doesn't, but that's not what Scrinzi is saying. The reason is doesn't is because we could work, for example, in Wigner quasiprobability representation: $$\rho\mapsto W(x,p) = \frac{1}{\pi\hbar}\int_{-\infty}^\infty\langle ...


7

For a general and brief overview of the mathematical framework of Quantum Mechanics, see this answer. In a nutshell, Hilbert spaces arise from the representation theory of C*-algebras, which are postulated to be the relevant mathematical object that describes a quantum theory (because it contains observables in its self-adjoint part, and states as special ...


5

If $A$ is self-adjoint, you can define $f(A)$ as a complex-valued observable, where $f: \mathbb R \to \mathbb C$ is a measurable complex-valued function: $$f(A) := \int_{\sigma(A)} f(x) dP^{(A)}(x)\:,$$ $P^{(A)}$ being the spectral measure (projector-valued) of $A$. $N= f(A)$ is a closed normal operator and admits a spectral decomposition $P^{(N)}$ ...


4

The general question is quite hard to tackle I think, because a rigorous motivation of Hilbert space would end up in the theory of operator algebras (see e.g. this answer) and the OP is probably not interested in these aspects at the moment. As for the example of spin, the Hilbert space in this case is still an $L^2$ space, but the functions are no longer ...


3

To specify a Gel'fand triple $(\Phi^*,\mathscr{H},\Phi)$ it is sufficient to specify the Hilbert space $\mathscr{H}$ and the topological vector space $\Phi\subset \mathscr{H}$. The necessary requirement is that the imbedding of $\Phi$ into $\mathscr{H}$ is continuous with respect to the topology of $\Phi$, so this gives the imbedding of ...


3

The time evolution of wave function is dictated by the Schrödinger equation, as you surely well know. Let's take the simple free particle in $\mathbb{R}^d$ (with mass $1/2$ and $\hbar=1$): $$i\partial_t \psi(t,x)=-\Delta_x \psi(t,x)\; ;$$ where $\Delta$ is the Laplacian operator (i.e. $\partial_x^2$). Mathematically, this is a linear PDE (partial ...


2

Comments to the question (v1): We will not discuss tachyonic states here, because they are pathological and signal an instability of the theory. Then $$\tag{1} p^{\pm}~\equiv~\frac{p^0 \pm p^1}{\sqrt{2}}~\geq~0 $$ is manifestly non-negative, since the energy $p^0\geq |p^1|$. In the light-cone formalism $p^{+}>0$ is strictly positive, since the special ...


2

Eigenvectors exist only for the point spectrum of an operator. For any other point of the spectrum one can only find a sequence of vectors for which $(A-\lambda I)u_n\to0$, where $A$ is said operator, and $\lambda$ is a point in the spectrum which is not an isolated point. So in this case there is a sequence of approximate eigenvectors. With a bit of extra ...


2

A self-adjoint operator $T$ on $L^2(\mathbb{R})$ has in its spectrum three different kinds of subspectra: A discrete point spectrum, a continuous spectrum, and a singular spectrum. The latter is physically discarded. The point spectrum consists of the eigenvalues of $T$, that is, the spectral values for which true eigenvectors in $L^2(\mathbb{R})$, and ...


2

In the position basis. $\Psi_n(x) = \langle \phi _n, \Psi \rangle \phi_n(x)$, and if the set of $\phi_n$ is a complete orthonormal set of functions, then: $$\Psi(x) =\sum_n \langle \phi _n, \Psi \rangle \phi_n(x)$$ where $\langle \cdot , \cdot \rangle$ is an inner product, and is not really Dirac notation. $\langle \cdot | \cdot \rangle$ is the same thing ...


2

Think of your functions just like vectors. You can add two (square integrable) functions and get another (square integrable) function. You can multiply a (square integrable) function by a scalar (pointwise) and get a (square integrable) function. So they add like vectors, they can be scaled like vectors. You even have a scalar product. So they are just ...


2

Let $$\tag{1} \hat{T}_{ik}~:=~\hat{n}_i \hat{n}_k-\frac{1}{3}\delta_{ik}\hat{\bf 1}.$$ The phrasing of the problem in Ref. 1 is indeed not the clearest, but by comparing with the given solution, it seems that Ref. 1 is performing a partial averaging over the Hilbert space of states with fixed value of the orbital angular momentum quantum number $\ell$ and ...


2

Item 1. has answered affirmative by Phoenix87 in the comments. Item 2. has been answered affirmative by yuggib in the other answer. The answers to item 3. and 4. have been given without proof by Phoenix87 in the comments, namely that the $\mathbb R$- and $\mathbb R^3$-Schwartz-space are isomorphic as rigged Hilbert spaces. A nice proof of this fact is given ...


1

You do not obtain the rules for the infinite-dimensional case by "proving" them from the finite-dimensional rules. Rather, you know that you need to have a Hilbert space, which is a complex vector space with an inner product, essentially. If you now search for infinite-dimensional Hilbert spaces that could possibly be used in quantum mechanics, you find ...


1

First of all, I would encourage you to think of position of acting on momentum states to the left, that is, to commute them with the bra: $$ ⟨\mathbf p|\hat x=-i\hbar \frac{\partial}{\partial p_x}⟨\mathbf p|, $$ where the differentiation is over anything to the right of it, so for instance $$⟨\mathbf p|\hat x|\psi⟩=-i\hbar \frac{\partial}{\partial ...


1

Nothing will help, a plane wave occupies all the space, and the mean value of the position doesn't make much sense, but in your case is zero, because the integrand in the 1st calculus is anti-symmetrical. But this, on one condition, namely if we write the integral as $lim _{a \to \infty} f^2(p_0) \int _a^a x d^3 x$ Otherwise it's hard to say what is the ...


1

The wave function only contains all the information about the system im so far as you consider it. Meaning each qualitatively different physical system needs its a modified Hilbert space to fit what can happen with the system. In case you have something like spin on its own in $ H_{Spin}$ and you want to look at a freely moving particle in $H_{free}$ that ...


1

If the operator is not self-adjoint then this is a possibility. If you search on phys.SE you will find questions about the expectation value of xp in the case of the QHO, and this turns out to be imaginary


1

No, the Hilbert space is not spanned by continuous "eigenfunctions" because they are not eigenfunctions at all! A self-adjoint operator $T$ on $L^2(\mathbb{R})$ has a point spectrum, a continuous spectrum and a singular spectrum. The latter is physically irrelevant. The point spectrum consists of the values $\lambda_i$ for which a true eigenvector ...


1

The way you've written the observable, let's call it $A$, indicates that $\vert a_n\rangle$ is a base of the Hilbert space you are acting on. Try to do the calculation: $$A \vert a_n \rangle = \sum_m a_m \vert a_m \rangle \langle a_m \vert a_n \rangle = \sum_m a_m \delta_{nm} \vert a_m \rangle= a_n \vert a_n \rangle $$ and understand what you get. Now ...


1

When you define a base of vectors, in your case $\{ |1,1\rangle, |1,0\rangle, |0,1\rangle, |0,0\rangle \} $ you assign to each vector some phase, e.g. the vector $|1,1\rangle$ has a certain phase that you don't mention explicitly outside the bra-kets. Now, in your calculi you may have superpositions of these vectors of the form $|\psi\rangle = ae^{i\alpha} ...



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