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7

This is a supplement to freude's correct answer: Hamiltonian is the infinitesimal generator of time translation defined as $$\mathrm{\hat{U}}(\mathrm dt)= 1- \frac{i}{\hbar} \mathrm{\hat{H}}(t)\mathrm dt\;.$$ Time-Evolution Operator: Let the system be at $|\phi\rangle\;.$ Now, let's wait for some time..... What is the probability amplitude of finding ...


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 ...


6

This is a scalar value that is a projection of the state $H|\psi \rangle$ on the state $|\phi \rangle$. The state $H|\psi \rangle$ results from the action of the operator $H$ on the state $|\psi \rangle$. If the state $|\psi \rangle$ is an eigenstate of the operator $H$, the expression can be rewritten as $E \langle\phi|\psi \rangle$. If the state $|\phi ...


4

This is because $H'=UHU^{-1}$ for a certain unitary operator $U$, therefore $\psi$ is an eigenvector of $H$ with an eigenvalue if an only if $U\psi$ is eigenvector of $H'$ with the same eigenvalue. Thus the two operators have the same point spectrum. $U = e^{i \lambda X/\hbar}$. From $[X,P]= i \hbar I$ one finds $$e^{-i \lambda X/\hbar}Pe^{i \lambda ...


3

$ \def\ee{\mathrm{e}} \def\ii{\mathrm{i}} \def\dd{\mathrm{d}} $ There are many ways to perform this calculation. Perhaps the simplest is to consider the object $a(t) = \ee^{\ii \omega_c t a^\dagger a } a \ee^{-\ii \omega_c t a^\dagger a }$ as the solution of the differential equation $$\dot{a}(t) = -\ii\omega_c a(t).$$ You can show that $a(t)$ satisfies ...


2

The answer is given by Prahar in his comments : \begin{equation} {\bf T} \cdot {\bf T}^\dagger= {\rm T}_{1}{\rm T}^\dagger_{1}+{\rm T}_{2}{\rm T}^\dagger_{2}+{\rm T}_{3}{\rm T}^\dagger_{3} \tag{01} \end{equation} For $k=1,2,3$ \begin{equation} {\rm T}_{k}{\rm T}^\dagger_{k}=\dfrac{1}{2\hbar}\left(\sqrt{m\omega}\ {\rm R}_{k}-\dfrac{i}{\sqrt{m\omega}} {\rm ...


2

It's simply the most general kind of interaction Hamiltonian you can write down in this simplified two-level system. On the 2D Hilbert space spanned by $\lvert R\rangle,\lvert L \rangle$, the most general linear operator is written as $$ A = a_\text{RR}\lvert R\rangle\langle R\rvert + a_\text{RL}\lvert R\rangle\langle L \rvert + a_\text{LR}\lvert ...


2

A wave function is an abstract mathematical function which could completely describe about the system under consideration. We define a wave function such that we could derive whatever information from it, provided that will not affect the state of the system. The wave function is not any operator. It's simply a function of position and time. Any ...


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

The precise statement of "self-adjoint operators generate continuous unitary symmetries" is Stone's theorem. It guarantees that there is a bijection between self-adjoint operators $O$ on a Hilbert space and unitary strongly continuous one-parameter groups $U(t)$ that is given by $O\mapsto \mathrm{e}^{\mathrm{i}tO}$. The definition of the exponential for an ...


2

When one writes $A=c$, where $A$ is an operator and $c$ is a number, it is implicit that the r.h.s. actually denotes $c$ times the identity.


2

General comments to the question (v1): Any textbook derivation of the correspondence between $$\tag{1} \text{Operator formalism}\qquad \longleftrightarrow \qquad \text{Path integral formalism}$$ is just a formal derivation, which discards contributions in the process, cf. e.g. this Phys.SE post. Rather than claiming complete understanding and existence ...


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 ...


1

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 ...


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

It's a case of bad labeling: the $i$,$j$ labels in Fig.1 and Eqs.(4-5) have different meaning. In addition, subscript 1 was dropped on all $B$'s in Eq.(5). Other than that, it's straightforward algebra: Start by rewriting the final result of Fig.(1) in the familiar operator-product form, expand, and rearrange: $$ \overline{\left[ E \cos(B_1\tau) - i {\hat ...


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( ...


1

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} ...


1

I) For the record, here is the operator calculation that OP wants to avoid. The benefit of the calculation is that the operators are not sandwich with any bra/ket representation, and hence we do not have to worry about whether the bra/ket representation is faithful. Let us put $\hbar=1$ for simplicity. The starting point is the CCR $$ [x^i, ...


1

Operators form a complete set if specifying eigenvalues of all of them already determines a state uniquely (up to normalization and phase). Since they commute, they will have simultaneous eigenspaces. Eigenstate is uniquely determined if all these eigenspaces are one dimensional.


1

I'm sorry I first misinterpreted your braces as anticommutators and not as set inclusions! First absorb the constants into the normalizations of H and B so the (1,1) entry of each is now 1. The common eigenvectors of them both are then the transposes of (1,0,0), (0,1,1), and (0,1,-1); the eigenvalues of each are, under H: 1,-1, -1 ; and under B: 1, 1, -1, ...


1

To quantize just means to impose this commutation relation: $$ [x_i , p_j ] = i \hbar \delta_{ij}.$$ By the way, remember that a function of position $f(x)$ may not commute with the momentum operator $p$. $A_i = A_i(x)$. Do $A_i$ and $p_j$ necessarily commute?


1

Expectation values of constants or numbers are just those constants or numbers. The expectation value of the anti commutator of $\hat x$ and $\hat p$, that is, $\langle\{\hat x,\ \hat p\}\rangle$, for the Harmonic Oscillator, or coherent states of the Harmonic Oscillator, is equal to $0$. $$\langle\{\hat x,\ \hat p\}\rangle = \langle\hat x \hat p + \hat p ...



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