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10

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 ...


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 ...


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 ...


5

The canonical quantization procedure prescribes that $$ [x,p] = \mathrm{i\hbar}$$ Now, take the trace on both sides. The trace of a commutator vanishes, and, if the Hilbert space $\mathcal{H}$ were finite-dimensional, we get $$ 0 = \mathrm{i}\hbar\mathrm{dim}(\mathcal{H}) $$ which is obviously false, so the assumption of finite dimensionality is wrong. ...


4

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 $\Psi$. ...


4

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 ...


4

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 ...


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 ...


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 ...


2

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 ...


2

The operator $d/dx$ isn't a "function" and Dirac surely never claims so. It's an operator, something that changes one function to another. By a function, we mean something that maps one number to another. Functions of $x$, like $f(x)$, may also be connected with operators on the space of (wave) functions. The wave function $\psi(x)$ is mapped to ...


2

You have to interpret $|\frac{d}{dq} \psi\rangle$. Knowing that decomposition of the basis $|q'\rangle$ gives : $$|\psi\rangle = \int dq' \psi(q') |q'\rangle \tag{1}$$ You have : $$|\frac{d}{dq}\psi\rangle = \int dq' \frac {d\psi(q')}{dq'} |q'\rangle\tag{2}$$ So, applying it to $|\psi\rangle = |q"\rangle = \int dq' \delta(q"-q') |q'\rangle$, you get : ...


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.


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 \begin{equation} |\psi\rangle = \alpha\,|0\rangle+\beta\,|1\rangle \end{equation} where the quantities $\alpha$ and ...


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 ...


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 ...


1

Your book is correct. Remember $ \psi $ is the wave function that represents the state of the system and is the solution of the Schrodinger's equation. Now, the Schrodinger's Equation is a linear partial differential equation (PDE) and the solution has several interesting properties: It has infinitely many particular solutions and we consider only those ...


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} ...


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 ...


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 ...


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 ...


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

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

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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