# Tag Info

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As other answers mention, it was originally (in QED) about getting a neutral vacuum. It is useful to go back to Schwinger's old version of QED, before Dyson's approach became accepted. See Pauli: Selected topics in field quantization. Pauli presents both ways of looking at it: 1) define the electric current as sum of two terms (p.20 [6.4]), such that the ...

3

Note that: \begin{aligned} \langle j | A \rangle & = \langle j |\left( \sum\limits_{i} a_i |i \rangle \right) \\& = \langle j| \left(a_1 |1 \rangle + a_2 |2 \rangle + a_3 |3 \rangle + \cdots + a_j |j \rangle + \cdots \right) \\& = a_1 \underbrace{\langle j | 1 \rangle}_{=\delta_{j1} = 0} + a_2 \underbrace{\langle j | 2 ...

2

If you think in terms of vectors and matrices with indices, $a_i$ is a vector and $\delta_{ij}$ is the unit matrix. The summation on the right hand side of your equation represents nothing but a multiplication of the vector $a_i$ with the latter. Since free indices are preserved, the remaining object carries the index $j$. You can write down an arbitrary ...

2

The choice of normal ordering prescription $:~:$ is typically adjusted to the choice of vacuum state $|\Omega\rangle$ so that the bra-ket-sandwich of normal-ordered operators $$\langle \Omega|:\hat{\cal O}_1\ldots \hat{\cal O}_n : |\Omega\rangle~=~0$$ vanishes. The relation of normal ordering prescription to Wick theorem and other operator ordering ...

2

In classical physics, quantities are ordinary, commuting $c$-numbers. The order in which we write terms in expressions is of no consequence. In quantum field theory (QFT), on the other hand, quantities are described by operators that, in general, don't commute. Classical physics is a low-energy approximation of quantum physics - the road from quantum to ...

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This is the same notation that you'll find in Weinberg's books. $$(\psi, \chi)$$ is the inner product of the two states $\psi$ and $\chi$, and corresponds to $$\langle \psi \mid \chi \rangle$$. So, the above corresponds literally to $$\frac{1}{\sqrt{\langle \psi_k \mid \psi_k \rangle}} \left| \psi_k \right>$$ This new object is just the normalized ...

4

Your intuition that If, instead, my measurement is only partly accurate and says that the momentum of the particle is in a set $\Delta =(a_x,b_x)\times(a_y,b_y)\times(a_z,b_z)$, will the measurement collapse the wave function into $P\Psi$ (where $P$ is the spectral projector of the momentum operator on the set $\Delta$)? is exactly correct. ...

1

The statement is simply false as it stands when adopting the standard Hilbert space formulation of QM. The true statement is that a self-adjoint operator with pure point spectrum admits a Hilbert basis made of eigenvectors. (It happens in particular, but not only, when either the operator is compact or its resolvent is.) The proof is not so simple and is a ...

0

“Always a polytope” – definitely not. Moreover, in certain situation $Ω$, if a closed set, may not change at all; I mean product with the 0-dimensional set of states $Ω_{\rm id} = \{1\}$ (one point), considered as a subset of 1-dimensional vector space $V_{\rm id} = {\mathbb R}$. It has the only effect, the unit effect, and corresponds to 1-state quantum ...

0

EDIT 2:There is no general procedure. It is not easy to find out how many commuting observables there are. In classical mechanics some systems are integrable. Then they have as many constants of motion as the number of degrees of freedom. For a system of N particles in d spatial dimensions times the number of degrees of freedom is Nxd. One particle and a ...

1

A physical state is defined by the density matrix, so, if you define the density matrix by : $\rho = \frac{|\psi\rangle \langle \psi|}{\langle \psi| \psi\rangle}$ it is easy to see that any multiplication by a complex number does not change the density matrix, so does not change the physical state. It is probably what Dirac means, while I have not the ...

2

Close, but not quite. Since quantum mechanics deals in probabilities, it is necessary to "normalize" the state in order to use it in later calculations. The most general state for the two-state quantum system you're considering would be $$|\psi\rangle = \alpha\,|0\rangle+\beta\,|1\rangle$$ where the quantities $\alpha$ and ...

2

Yes, that is correct. A more general form of the superposition of the stationary state would be $$a|0\rangle + b|1\rangle$$ where $a,b$ describes the probability of each state. The state : $$|0\rangle + |1\rangle$$ assumes that the state $|0\rangle$ and $|1\rangle$ are equally probable.

1

Stated in a simpler way, kets are the generalisation of vectors to complex and potentially continuous/infinite dimension space (Hilbert spaces). Yet you can keep in mind the image of a vector to begin with. When you multiply a vector by a nonzero real number, its direction does not change. The same is valid for vectors from a Hilbert space. If you multiply ...

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Obviously I cannot know what Dirac is thinking, but I think it is just that his direction does not correspond exactly to your direction. We "know", just as Dirac does, that quantum states are members of some Hilbert space $\mathcal{H}$. We also know that scalar multiplication should not change the state, so $\lvert \psi \rangle$ and $\lambda \lvert \psi ... 1 Hints to the question (v2): First note that the operator norm$||A||=||UA||=||AU||$of an operator$A$is invariant if we compose with an unitary operator$U$from either left or right. Therefore$\dot{\rho}(t)$is not the zero-operator:$|| \dot{\rho}(t) || = || [H, \rho(0) || \neq 0. $1 The superposition principle of quantum mechanics is not destroyed by quantum (hamiltonian) unitary evolution operator$U(t_0,t) = \mathrm{e}^{\mathrm{i}H (t - t_0)}$as per @ACuriousMind's answer. Event if you dont know of the evolution operator in terms of the hamiltonian (which can be derived easily from the Schrodiger equation), still the fact that the ... 4 The time evolution operator of a quantum system is (in units with$\hbar = 1$) $$U(t_0,t) = \mathrm{e}^{\mathrm{i}H (t - t_0)}$$ and the "stationary states" are the eigenstates of this operator, i.e. eigenstates of the Hamiltonian. If you are given a collection of stationary states (not a basis of the space, mind you)$\{\lvert \psi_E \rangle\}$with$H ...

5

It is possible indeed !! It is called Hilbert Schmidt scalar product, it is defined in a Hilbert space of bounded compact operators including trace class operators. $$\langle A|B\rangle := tr(A^\dagger B)\:.$$ The space of Hilbert Schmidt operators is made of all bounded operators $A$ in the considered Hilbert space, such that $A^\dagger A$ is trace ...

1

Suppose your initial state is $\lvert 2\rangle$ and that the states $\lvert 0 \rangle$ and $\lvert 1 \rangle$ have lower energies than $\lvert 2 \rangle$. Assuming that there is no so called selection rule that prevents $\lvert 2 \rangle$ from emitting a photon and end up in $\lvert 0 \rangle$ or $\lvert 1 \rangle$, then the final state will be $$\lvert 2 ... 8 The expectation value of energy is something else than the energy in a particular experiment. With your choice of the initial states, the photons emitted (negative difference) or absorbed (positive difference) will have energies either$$ E_1-E_0 \text{ or } E_2-E_0 \text{ or } E_1-E_5 \text{ or } E_2-E_5 $$If each of the four transitions were equally ... 1 As you stated already, a measurement of the energy of the hydrogen atom must return an energy eigenvalue. Measuring before and after a transition gives us two energies E_n and E_m. This is always true, regardless of the fact that the expectation value of the energy before measuring might not be a difference of E_p and E_q for some p,q: The actual ... 2 The key concept to look for is displaced number states. These are, quite simply, the number states |n⟩, moved by the displacement operator$$D(\alpha)=\exp\left[\alpha a^\dagger-\alpha^*a\right]$$to some point \alpha=x+ip on the complex phase space. The ground state of a harmonic oscillator which has been displaced to a real \alpha=x is, as you ... 1 I) Right, the operator$$\tag{1} \hat{A}~\equiv~ \hat{a}-\alpha{\bf 1}, \qquad \alpha\in \mathbb{C},$$satisfies the same commutation relations$$\tag{2} [\hat{A},\hat{A}^{\dagger}] ~=~{\bf 1}$$as$$\tag{3} [\hat{a},\hat{a}^{\dagger}] ~=~{\bf 1}.$$(In OP's example the complex number \alpha=-1.) II) Define number operator$$\tag{4} ...

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I have 5 bags labelled 1 to 5, and I have randomly dropped the letters A to J into the bags. You choose a letter at random and you win as many Francs as the number on the bag containing your letter. If I have distributed the letters evenly, then there should be 2 letters in each bag, so we could say that ψ(bagnumber) = ψ = sqrt(2). But if we want ...

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The expectation value (of position) represents the average value (position) for the particle (it has units of length in this case) which is different from the actual location of the particle (also units of length). For example, take an electron on a hydrogen atom; the expectation value for all energy levels is at the nucleus even though many of the energy ...

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In position-space (that is, when your functions are functions of x), the function $\int|\Psi|^2$ gives the probability of finding the particle in a given range. The expectation value of x is where you'd expect to find the particle. It is often essentially the weighted average of all the positions where the probability density, $|\Psi|^2$, is the weighting ...

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Let $\Omega\subseteq \mathbb{R}^n$; then $\int_\Omega \lvert\psi(x)\rvert^2dx$, for a normalized function $\psi\in L^2(\mathbb{R}^n)$ gives the probability that the particle is in the region of space $\Omega$, but does not give any further information on its position. If you want to obtain a quantitative information on the latter (within the limits of ...

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Expectation value is a different concept from probability. In fact, you can have an expectation value of energy, angular momentum, etc., not just for position. An expectation value of an observable for a given state $\Psi$ is the average value of a large number of measurements of that observable, assuming each measurement is made on the same state $\Psi$. ...

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