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4

$\hat{X} = i\partial_{k_x}$ in momentum space is wrong. The left hand side is an operator, the right hand side is the representation of an operator when acting on functions, taken with respect to the scalar product with the basis $|\ p\rangle$. In particular we have $$\langle x\,|\,\hat{x}\,|\,\phi\rangle = x\phi(x)$$ and $$\langle p\,|\,\hat{x}\,|\,\... 2 You have to keep in mind that an integral is (more or less) a sum, and therefore in a double or multiple integral, you have cross terms that have to be correctly time-ordered. The usual Riemann integral does not take in account the non-commuting nature of its integrand, so you have to make it manifest with the time-ordering symbol \mathcal{T}. Imagine for ... 3 Consider$$\int_0^t dt'\int_0^{t}dt'' \mathcal{T}\left(H(t')H(t'')\right)\begin{align} &= \int_0^t dt'\int_0^{t'}dt'' \mathcal{T}\left(H(t')H(t'')\right) \\ & \qquad +\int_0^t dt'\int_{t'}^{t}dt'' \mathcal{T}\left(H(t')H(t'')\right) \\&= \int_0^t dt'\int_0^{t'}dt'' H(t')H(t'') \\ & \qquad +\int_0^t dt'\int_{t'}^{t}dt''H(t'')H(t') \end{... 4 Time-ordering is needed if the Hamiltonians H(t^{\prime}) and H(t^{\prime\prime}) at different times do not commute. Example: If the Hamiltonian isH(t) ~=~ \left\{\begin{array}{ccl} \color{Red}{\diamondsuit}&\text{if}& t<0, \cr \clubsuit &\text{if}& t>0, \end{array} \right.$$then the non-time-ordered product (\int\! dt ~H(t)... 1 I will answer with an example on how eigen functions of momentum do not exist in a Hilbert space in general. When they do not, we say the momentum operator is not self-adjoint. Consider a one dimensional infinite potential well. Lets us place the walls at x=0 and x=L. For x\ge L and x\le 0, the potential V(x)=\infty and therefore we put the ... 3 The boundary conditions determine whether an operator is Hermitian or not. Once you know your operator is Hermitian, you have the results on the spectrum. Without boundary conditions the momentum operator need not be Hermitian, hence its spectrum can have non-real values (here I am assuming that by Hermitian you actually mean self-adjoint). As an example, ... 4 If all truncated n-point functions vanish for n>2 (i.e. we are dealing with a so-called generalized free field), microcausality for vacuum expectation values and at the operator level are equivalent. The former, on its turn, follows from Lorentz invariance alone in the case of scalar (but not necessarily free) fields, as shown by Pierre-Denis Methée, ... 2 Comments to the question (v3): The time-ordered product of operators$$T[A_1(t_1)\ldots A_n(t_n)] ~:=~\sum_{\pi\in S_n} \theta \left(t_{\pi(1)} > \ldots > t_{\pi(n)} \right) (-1)^{\varepsilon_{\pi,A}} A_{\pi(1)}(t_{\pi(1)})\ldots A_{\pi(n)}(t_{\pi(n)}) \tag{1}$$is graded symmetric. [Here (-1)^{\varepsilon_{\pi,A}} is a sign factor in the case ... 1 I'm not sure what "Pauli's argument" precisely is because he refers to pages in the first edition of Dirac's Principles of Quantum Mechanics which contain nothing of evident relevance in my fourth edition, but the more common thing to say is that it is the boundedness of the energy from below that forbids a naive time operator, not the discreteness. However,... 0 Commutation relations between position and momentum operators, in quantum mechanics, are valid only when the actions of either of the two operators is contained in the domain of the remaining one. In particular we have:$$ [x,p]\psi = (xp)\psi-(px)\psi. $$In order for the above to be defined \psi must be in the domain of definition of both operators and, ... 2 For the schrodinger picture if you take a state |\psi(t)\rangle satisfying the schrodinger equation i\hbar\frac{\partial }{\partial t}|\psi(t)\rangle=H|\psi(t)\rangle, and write a time evolution operator U such that |\psi(t)\rangle=U(t,t_0)|\psi(t_0)\rangle, then this gives an operator equation that U must satisfy namely:$$i\hbar\frac{\partial }{\...

2

The extra one simply reflects the commutation relationship $[a,\,a^\dagger] = \mathrm{id}$ of the quantum mechanical harmonic oscillator. If each annihilation operator in the expression for $E^+$ acts only on its corresponding mode, and likewise for the creation operators in $E^-$, then the action of $E^-\,E^+ -E^+\,E^-$ on a state involving only excitations ...

0

$\chi_I$ is a number, not an operator, do not mix up $\chi_I$ with the projection operator. This whole formalism is actually very simple in this case. $P_I$ is a projection operator onto the space of states which correspond to a particle being inside the interval $I$. Suppose we live in discrete space for the time being so we can work with explicit ...

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Notice that the probability of measuring say the position of a particle whose wavefuction is $\psi(x)$ in the interval $I=(a,b)$ is $$\int_a^b \left|\psi(x)\right|^2 \mathrm{d}x.$$ We can define a multiplication operator on the state space much like the position operator $\hat{X}\psi(x)=x\psi(x)$ as follows. $P_I(\psi)=\chi_I(x)\psi(x)$. It is a projection ...

0

For any self-adjoint operator $A$ (with continuous or discrete spectrum), the so-called spectral theorem tells you there is unique a family $(P_{\lambda}(A))_{\lambda\in\mathbb{R}}$ of orthogonal projections (called spectral family) such that $P_{\lambda_2}(A)-P_{\lambda_1}(A)$ gives you the projection on the subspace with spectrum in the interval $[\... 2 The naive thing to do would be to just replace$x$by$\hat{x}$and$p$by$-i\hbar \nabla$. Note however that$(\hat{x}\hat{p})^{\dagger} = \hat{p}\hat{x} \neq \hat{x}\hat{p}$, i.e. the operator is not Hermitian due to the noncommutativity of$\hat{x}$and$\hat{p}$, which is a problem of course. So what can we do to fix this? One possible way is to ... 3 For a (sufficiently nice) expression$f(X)$, where$X$ranges over a vector space), the directional derivative$Ydf(X)$with respect to$Y$(in the same vector space) is the coefficient of$\epsilon$in an expansion$f(X+\epsilon Y)-f(X)$where$\epsilon$is a formal variable with$\epsilon^2=0$. The functional derivative$\delta f(x)/\delta(X)$is the ... 2 The angular momentum is$\vec L~=~\vec r\times\vec p$which according to a bulk system with a moment of inertial$I$is also$\vec L~=~\hat n I\omega$. Here the unit vector$\hat n$is normal to the plane of$\vec r$and$\vec p$. In Hamiltonian mechanics we have $${\dot L}_i~=~I\dot\omega_i~=~\{H,~L_i\}_{pb}~=~0,$$ for$i$the coordinate direction which ... 8 There doesn't exist any procedure to uniquely associate a Hermitian operator$L$to a function of the phase space$f(x,p)\$. Quantum mechanics is a theory that exists independently of classical physics. Quantum mechanics is not just a cherry on a classical pie that needs the classical theory to exist at every moment. If we want to define a quantum theory, we ...

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