# Tag Info

## Hot answers tagged hilbert-space

12

Yes, you just have to check that $e^{-i\beta P/\hbar}|q\rangle$ is an eigenket of $Q$ (uppercase represents operators) with eigenvalue $q+\beta$. We evaluate $$Q \big ( e^{-i\beta P/\hbar}|q\rangle \big )$$ by using the first identity (which is actually a consequence of the canonical commutation relation): $$Q e^{-i\beta P/\hbar} = e^{-i\beta P/\hbar}Q + ... 4 It measures the (square of the) magnitude of the spin angular momentum, ignoring the spin’s direction. 4 Are -|0\rangle, i|0\rangle, \frac{1+i}{\sqrt2}|0\rangle valid (that is, physically possible) kets? How are they different from |0\rangle? Yes, they are valid, but describe the same physical state. A state can be characterized by its expectation values, and a phase factor e^{i\varphi} cancels with the complex-conjugate factor from the bra. ... 2 OP asks good questions. Let us try to sketch the logic of the LSZ reduction formula. In the Heisenberg picture, the real field \varphi(x) has a Fourier expansion$$ \varphi(x)~=~\int \widetilde{dk}\left[a({\bf k})e^{ik\cdot x}+ a^{\dagger}({\bf k})e^{-ik\cdot x} \right], \tag{3.19} $$where k\cdot x= {\bf k}\cdot {\bf x}-\omega_{\bf k}t and \omega_{... 2 The answer depends on which type of "Majorana operator" you are considering. If you mean a relativistic four-component Majorana fermion field \Psi (which in the absence of interactions solves the Dirac equation), then it is \Psi^\dagger \neq \Psi. Instead the correct statement is that \Psi^C := \mathcal{C} \overline{\Psi}^T := \mathcal{C} (\Psi^\dagger ... 2 If you want X have eigenvalue x_0 at t_0 you should take either t_0=0 or take U(t)= \exp\{-iH(t-t_0)\} so that U(t)={\rm Id} at t=t_0. 2 In order to have dissipation you need somewhere for the energy to go, and this somewhere has to be included in the quantization process. In addition to absorbing the energy, the extra system causes quantum decoherence and so makes the problem quite tricky. The damped harmonic oscillator is one of the problems that were attacked and solved by Caldeira and ... 2 Yes this is the right way to look at this except you need U(s) instead of U(s-\epsilon) in your formula. If you have a map s\mapsto U(s) into the set of bounded operators on a Hilbert (e.g., unitary operators), then you have two main approaches for defining the derivative$$ \frac{d}{ds}U(s)=\lim_{\epsilon\rightarrow 0} \frac{1}{\epsilon}(U(s+\epsilon)-...

2

For a particle to have a definite energy, it must be in an energy eigenstate of the Hamiltonian that describes it. But this in no way means that our system always has to be in one of the energy eigenstates. Take the simple case of particle in a box. The only condition on the wavefunctions is that it is $0$ at and outside the boundaries of the box and that ...

2

No, take e.g. $X=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $A_0=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ (only one $A_i$).

2

In the traditional formulation of Quantum Mechanics, it is stated that observables (which are physical quantities, like energy, position, momentum etc.) are represented by operators with such and such properties. This is the proper wording, "are represented by". Measurements and their link to theory have to do with the vast subject of "Interpretations of ...

1

Suppose that we have a state, which we can describe as a linear combination in some basis $|\phi\rangle = \sum_m c_m |\psi_m\rangle$ with $||\psi_m||=1$ for all $m$. Acting with the projection $P_n = |\psi_n\rangle\langle\psi_n|$ will give as exactly $c_m |\psi_n\rangle$, which is what a projection should do. The resulting state lies completely in the ...

1

You just need to keep scrolling down in the Wikipedia article, I will just add the mention of the Wick rotation. First of all this is a hypothesis, so you can understand it as a "well motivated guess", the reasoning coming from (as stated in the Wiki): $$\bar{A} = \lim_{T\rightarrow \infty} \frac{1}{T} \int_0^T {\rm d}t\, \langle \psi(t)|\hat{A}|\psi(t)\... 1 First, the one-particle state localized in the point, e.g. \phi(x)|0\rangle is very singular. What behaves more nicely insteas is a smeared state \int_V d^D x \eta(x) \phi(x)|0\rangle. However if we look at this "particle at a point" we will see,$$ \phi(0)=\int \frac{d^{D-1}p}{(2\pi)^D\sqrt{2 E_p}} a^\dagger(\vec{p})|0\rangle $$I.e. to produce ideally ... 1 If we have a rotating body, it's classical rotational energy is$$ E_{rot} = \frac{L^2}{2\theta}  Therefore, if we include a rotational particle in quantum mechanics it seams "natural" to use the same term and replace $L^2$ by it's operator -- the mathematical argument involves some group theory and so called generators [see e.g. the book of Sakurai for a ...

1

The term "slogan" may not be the best choice of words here. This is not about a commercial for toothpaste. Also, taking as a starting point "constructing a QFT means constructing a Fock space", while to some extent true in the free case, is not correct in general. If you are familiar with the Wightman axioms defining a QFT and want to understand what they ...

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