Tag Info

New answers tagged

0

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 ...


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

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} ...


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 ...


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 ...


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 ...


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

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 $\Psi$. ...


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

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 ...


0

Quantum states are not always complex and infinite dimensional. For example state which represents only spin of spin-1/2 particle is 2-dimensional and in matrix representation can be written for example as (1,0) for "spin up" and (0,1) for "spin down". They are complex because in general instead of ones and zeros there can be any complex numbers (up to ...


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. ...


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

Consider a system in the state $$|\psi\rangle=\sum_kc_k|\psi_k\rangle$$ Suppose that this system couples to another system. The coupling will leave unchanged some set S of commuting observables of the original system and will change the others. In general, unless the system happens to be in an eigenstate of that observable this will change the expectation ...


1

You can find derivation of these operators in most standard quantum mechanics textbooks. For your convenience, see https://en.wikipedia.org/wiki/Momentum_operator and https://en.wikipedia.org/wiki/Position_operator. For the second question, Paul Dirac said in his classic The Principles of Quantum Mechanics: A measurement always causes the system to jump ...


1

Let $A$ hermitian operator corresponding to an observable. If $\psi$ is an eigenfunction of A, then $$ A\psi = \lambda\psi $$ We say: $A$ has the value $\lambda$ on $\psi$. If it's not an eigenfunction, then $$ A\psi = \lambda_1\psi_1 + \lambda_2\psi_2+\dots $$ That is, after the measurement the state changes and you cannot associate a definite value ...


2

The former and the latter are really the same: "$c_n=\psi(x)$". If you want to measure positions, then possible outcome states are $|x\rangle$, therefore you write $$ |\psi\rangle = \sum_x|x\rangle\langle x|\psi\rangle:=\sum_x\psi(x)|x\rangle :=\sum_xc_x|x\rangle $$ This tells you, the probability to find the particle at position $x$, i.e. to measure it in ...


2

This wavefunction is an idealization of a wave function that has very very steep, but not discontinuous, behavior at $0$ and $a/2$. You are right, the wavefunction as written is not a proper wave function. That's a good observation. Often in physics the math is much easier if a real situation is modeled by one that is close to it, but mathematically more ...


2

$$ <\hat{A}> = \int \psi^*(x)\hat{A}\psi(x) dx $$ now $\hat{A}\psi(x)=a(x)\psi(x)$ so, $$<\hat{A}>=\int a(x)|\psi(x)|^2dx$$ Let $|\psi(x)|^2dx = d \mu$ now $\int|\psi(x)|^2dx=\int d\mu = 1$ $$ <\hat{A}> = \int a(\mu)d\mu $$


0

Let us use bra-ket notation. Suppose that the operator $\hat A$ has discrete and bounded eigenvalues $\mathcal A =\{A_1, A_2, \ldots\}$ with eigenkets $|A_1\rangle, A_2\rangle,\ldots$. The eigenkets form a complete orthogonal set since $\hat A$ is symmetric. Then any ket $|\psi\rangle$ can be expanded, $$|\psi\rangle = \sum \psi_n |A_n\rangle.$$ Since the ...


0

Observables are associated with linear operators, not measureable functions, so how can we talk about the expectation of a linear operator? The "expectation of a linear operator" is a term from quantum theory. It is defined by the integral $$ \int \psi^*(x)\hat{A}\psi(x) dx $$ or similar. The meaning of this number is not necessarily the same as ...


1

There are two answers to this. One answer simply points out that the probability of the jth outcome specified by the Born rule $p_j = tr(\rho\hat{P}_j)$, where $\hat{P}_j$ is the projector onto the jth outcome, satisfy the axioms of probability: http://mathworld.wolfram.com/ProbabilityAxioms.html. Another answer is that the Born rule can be explained ...


2

Since you want a bit of mathematical rigor: A quantum state is a self-adjoint positive trace class operator on a Hilbert space with trace 1. This is called density matrix $\rho$. In its simplest form, given $\psi\in \mathscr{H}$, $\rho$ is the orthogonal projector on the subspace spanned by $\psi$. Let $E_\rho(\cdot):D_\rho\subset\mathcal{A}(\mathscr{H})\to ...


2

The quantum state and the state vector is exactly the same thing. The word "vector" is meant to emphasize that quantum states form a vector space, the so-called Hilbert space. Because they're synonyma, the question in the first paragraph is a meaningless talkative tautology, like "Is a car meant to denote just one vehicle or are many cars meant to be many ...


1

You are maybe making confusion between the action of an observable (operator), and the measurement process. In particular: $A\psi$ is simply a vector of the Hilbert space. In my opinion it has not much sense of talking about "initial" and "terminal" state because you are not looking at a dynamical situation. If you want to know the average value of an ...


1

Regarding the first part of your question,they have just inserted a complete set of basis because $|\phi>$ is a basis in some infinite dimensional Hilbert space (in your case), therefore sum (integral) of all such bases is identity on the Hilbert space. Note that in second part $\langle r|\phi\rangle$=$\phi(r)$.


1

So first of all, the first equation you gave is only correct, if the $|ø\rangle$ form a basis. It has nothing to do with "in which basis they are". The easiest way to understand this is probably with a 3D vector-analogy. So if $b_i$, $i=1\dots3$ form a basis, for any vector $v$ it is legitimate to write $$v=\sum_{i=1}^3 b_i (b_i\cdot v)$$ There, the ...


2

No $\hat\phi|0\rangle$ is not an eigenvector of $\hat\phi$. You can see this, for example, by writing out $\hat\phi$ in terms of creation and annihilation operators, then compare $\hat\phi|0\rangle$ against $\hat\phi^2|0\rangle$, and observe that one is not a scalar multiple of the other. So as you suspected, eq. 5 is not correct To obtain some analogy of ...


3

You can utilize the construction of the $Q$ space, as described in Reed and Simon vol.2, page 228-230. Oversimplifying, you can make the analogy $\lvert \phi\rangle \sim \lvert x\rangle$, but the associated momentum is not $\hat{p}$, but $\hat{\pi}$ (the canonical conjugate momentum of the field $\hat{\phi}$). With slightly more precision: the Fock space ...



Top 50 recent answers are included