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10

This is indeed possible only in some situations, e.g. when the continuous spectrum is absent (it may also consist of a single point, see Valter Moretti's comment below). A sufficient condition for that to be true is that either the Hamiltonian is compact or it has compact resolvent. Sadly, very few interesting Hamiltonians satisfy that property (an example ...


6

Take your first relation for the 3 Pauli matrices individually: $$\sigma_1(t)=U^\dagger\sigma_1(0)U$$ $$\sigma_1(t)=U^\dagger\sigma_2(0)U$$ $$\sigma_3(t)=U^\dagger\sigma_3(0)U$$ Now you define a "vector" for notational convenience like the OP says in the question. I will choose to rewrite it as a column vector to visually show the relation of the above 3 ...


5

For the absolutely continuous part of the spectrum of a self-adjoint operator $H$, the "density of states" is provided by the Radon-Nikódym derivative of the spectral measure of $HP_{ac}$ with respect to Lebesgue measure, where $P_{ac}$ is the orthogonal projection onto the absolutely continuous subspace of the domain of $H$. This formula is well defined ...


4

Heisenberg's uncertainty principle is in fact not a principle but a consequence of the operator formalism of QM. If we associate to the operator $X$ the standard deviation $$\Delta_X = \sqrt{ \langle{X^2}\rangle -\langle X \rangle^2}$$ it can be then shown that, given two operators $A,B$ $$\Delta_A \Delta_B \geq \frac{1}{2} \left| \langle ...


4

Historically, you probably want to start with the de Broglie relations (i.e. $p = \hbar k$), which are just a wild guess. This immediately pops out the form of $p$ as an operator if the wavefunction is a plane wave. Mathematically, $p$ should be defined as the generator of translations (or equivalently the conserved quantity corresponding to translational ...


3

I am not sure I understand perfectly you question but formally in the canonical ensemble we can write the partition function $Q(\beta)$ as being: \begin{equation} Q(\beta) = \int\cdot \cdot \int d\mu(x)\: e^{-\beta H(x)} = \int_0^{+\infty} dE \: \rho(E)e^{-\beta E} \end{equation} where $d\mu(x)$ is the volume measure for the micro states in the system, ...


3

Let us assume that the operators are normalized so that $a^\dagger a$ has eigenvalues $0,1,2,3,\dots $. The prefactor in the exponent is $\mu h/2\pi=\mu\hbar$ must therefore be dimensionless, too. So $\mu$ has to have the units of the inverse action. Now, if $a=(x+ip)/ \sqrt 2$ – let's simplify the "masses etc." and set $\hbar=1$, then $$a^{\dagger 2} - a^2 ...


3

Similarly to AccidentalFourierTransform I am not sure to understand well your issue. However there is a crucial missed point in your argument, usually absent in many textbooks on these topics. It is true that decomposing $H$ as $H= \hbar\omega( a^\dagger a + \frac{1}{2})$ and taking the relations (following from CCR) $[a, a^\dagger] =I$ into account one ...


3

Well, I'm not sure I understood your question so I'm going to write what I think and let's see if it's useful :-) The algebra $[a,a^\dagger]=1$ is all you need to diagonalise $H$, but this is because what $H$ looks like: $$ H=\omega a^\dagger a $$ The important observables, namely $H,P,X$, can be written as polynomials in $a,a^\dagger$: \begin{aligned} ...


3

It's not true. For example take $$ A = \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} \qquad\qquad\qquad B = \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix} $$ you have $A^2=0$ so that $[A^2,B]=0$, but $$ [A,B] = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}. $$ You must add some condition, for example if you know also $$ [A,[A,B]]=0 $$ ...


3

The annihilation operator is a linear operator. A linear operator can be applied to ANY state. And yes, it returns zero when applied to the ground state. You can really take this as a definition. The definition of the momentum operator $\hat{p}$ is the operator such that $\langle x|\hat{p}|\psi\rangle=-i\hbar \psi'(x)$. One could write this as $\langle ...


3

If $|\pm\rangle$ are the eigenvectors of ${\hat \sigma}_x$, ${\hat \sigma}_x |\pm\rangle = \pm |\pm\rangle$, then a rotating $x$-basis is defined as $$ |+\rangle(t) = \exp\left(-i\frac{\omega t}{2} {\hat \sigma}_x\right)|+\rangle = e^{-i\omega t/2 } |+\rangle $$ $$ |-\rangle(t) = \exp\left(-i\frac{\omega t}{2} {\hat \sigma}_x\right)|-\rangle = e^{i\omega t/2 ...


2

Everything you write is correct and there is no inconsistency. When you write $\left<x\right|Hf(x)g(x′)\left|x′\right>$ you just, indeed, put the numbers on the right. But numbers commute so there is nothign wrong with it. Note that you have really no conclusion from it. You can't even transform back to operators on the form ...


2

It's because always even number of Fermionic fields appear in the Hamiltonian. for example the Dirac Lagrangian for free electron: $\mathcal{L}=i\hbar\bar \psi(\gamma^\mu\partial_\mu-m)\psi$ has two $\psi$s. Invariance of the theory upon a global gauge transformation requires that each term in the Lagrangian have an even number of Fermionic fields. Since an ...


2

If $\mathbf{x}=\left(x_{1},x_{2},x_{3}\right)$ is a 3-vector rotated to $\mathbf{x}^{\prime}=\left(x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime}\right)$ then this rotation is expressed via special unitary matrices $U \in SU\left(2\right)$ as follows : \begin{equation} \mathbf{X}^{\prime}\equiv \begin{bmatrix} ...


2

Short answer to your question: yes. There's no reason for the path integral and the operator formalism to be equivalent. A simple example are all the non-Lagrangian theories. We know (some of) them through their operator algebra, but there's no corresponding Lagrangian. This might sound strange to most people, like myself not long ago, but it is not hard ...


2

Assume that $g$ is a function in the space of rapidly decreasing functions with all their derivatives of any order, the so-called Schwartz space ${\cal S}(\mathbb R^4)$. Define $$\hat{g}(k) = \frac{1}{(2\pi)^2} \int_{\mathbb R^4} e^{-i k_\mu x^\mu} g(x) d^4x\::$$ As a consequence, as is well known, $\hat{g} \in {\cal S}(\mathbb R^4)$ and $$g(x) = ...


2

PART 1 : Taylor Expansion ( see @Prahar comment ). In the following : $$ \mathbf{x}=x^{\mu}, \quad \mathbf{k}=k_{\mu}, \quad \partial_{\mu}=\dfrac{\partial}{\partial x^{\mu}} , \quad \mu=0,1,2,3 \tag{1-01} $$ $$ \mathbf{k}\cdot\mathbf{x}=k_{\mu}x^{\mu}=k^{\mu}x_{\mu}, \quad \text{Einstein's convention on}\: \mu \tag{1-02} $$ Suppose now that the ...


2

Yes, it is. The definition of the exponential of an operator is $$\exp(\hat X) = \sum_{k=0}^{\infty} \frac{\hat X^k}{k!} = \hat1+\hat X+\frac{\hat X^2}{2}+\dots$$ where $\hat 1$ is the identity operator. So if you stop at the first order you will have indeed $$\exp(\hat A+\hat B) = \hat 1 + (\hat A + \hat B) $$


2

The point is that the domain $D(P)$ of $P$ must be such that $P$ is (essentially) self-adjoint thereon. Otherwise it does not represent an observable. I am assuming that $D(X)= L^2([0,L],dx)$ instead, where $X$ is automatically self-adjoint. The vector $\psi$ you use to prove Heisenberg inequality has to belong to $D(PX) \cap D(XP)$ as you see by direct ...


2

The most effective way to think about such problems is by means of operator algebras. The best way to treat systematically quantum observables (and their associated unitary operators) is to collect them in a C$^*$-algebra, roughly speaking a Banach algebra where observables have a norm, can be summed and multiplied (with some additional technical ...


2

If you ask about what plays the role of the state of the system at some given time than the answer is: nothing. You talk only about the initial state and what you get in the measurements (expectation values or probabilities of outcomes for observables). The Heisenberg picture is very "Copenhagen" in its spirit and abstracts itself from what's happening with ...


2

What's special about $H$ is that it generates time translations. This means that operators evolve, by postulate, through Heisenberg Equations of motion $$ i\frac{\mathrm d}{\mathrm dt}\mathcal O(t)=[H,\mathcal O] $$ modulo an explicit time dependence. Therefore, if an operator commutes with the Hamiltonian, $[H,\mathcal O]=0$ we automatically conclude that ...


1

Momentum and position are conjugate variables in classical mechanucs, which means they satisfy the Poisson bracket relationship. When quantum mechanics was invented the Poison bracket relation was replaced by the operator commutation relationship which results in the relation under consideration.


1

Here, $V_0(k_2)$ could have been replaced by $e^{k_2 \alpha_{-1}} e^{-k_2 \alpha_1}$ while the factors from $\alpha_{\pm n}$ for $n\gt 1$ could have been neglected because $\alpha_n$ annihilates everything that appears in the matrix element on the right side from $V_0$ (because it ultimately annihilates $|0\rangle$), and similarly for $\alpha_{-n}$ that ...


1

The physical states in the Heisenberg picture are frozen in time, and can be made to coincide with the Schrodinger-picture state at any given time $t_0$. In other words, $$|\psi(t_0)\rangle^S=|\psi\rangle^H$$ which doesn't evolve with time.


1

The adjoint of an operator is obtained by taking the complex conjugate of the operator followed by transposing it. i.e., $(A)^\dagger_{ij}=\left((A)^T_{ij}\right)^*=\left((A_{ij})^*\right)^T=A_{ji}^*$ You can do it in any order. The adjoint of an operator is the infinite dimensional generalization of conjugate transpose, where you find the transpose of ...


1

First, let's clarify the expression of $|\phi\rangle$. The kets $|+\rangle_z$ and $|-\rangle_z$ are eigenvectors of $\hat{S}_z$ such that $$ \hat{S}_z |+\rangle_z = +\frac{\hbar}{2}|+\rangle_z \\ \hat{S}_z |-\rangle_z = -\frac{\hbar}{2}|-\rangle_z $$ This means that in the $\{|+\rangle_z = ...


1

The answer is no, and the details are clearly spelled out in Glauber's Les Houches lectures (circa 1964). Glauber introduces a "T-representation" which can represent any operator in the Fock space of harmonic oscillator states, a less general "R-representation" which can represent any density operator, and the still less general "P-representation" which can ...


1

Hints: The starting point is the 2-point relation $$T(\phi(x)\phi(y)) ~-~:\phi(x)\phi(y): ~=~ C(x,y)~{\bf 1}, \qquad C(x,y)~\equiv~\langle 0 | T(\phi(x)\phi(y))|0\rangle,\tag{1} $$ cf. this Phys.SE post. The relevant Wick's theorem is a nested Wick's theorem $$ T(:\phi(x)^n::\phi(y)^m:)~=~\exp\left( ...



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