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Shankar is being a little bit sloppy with the term "state," although not that sloppy since he does use the qualifiers "quantum" and "classical" to achieve clarity. In classical mechanics, the position of a particle is represented by a point $\mathbf r$ in three dimensions $\mathbb R^3$. To rotate the configuration of the classical particle, then, one would ...

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If we are in the Schrödinger picture, we have $$i\hbar\frac{d}{dt}\psi(t)=H_S(t)\psi(t), \qquad\psi(0)=\psi_0$$ Then, we can describe the time evolution via a unitary propagator $U(t,0)$, such that $$\psi(t)=U(t,0)\psi_0$$ Substituting, leads to \begin{gather}&i\hbar\frac{d}{dt}U(t,0)\psi_0=H_S(t)U(t,0)\psi_0&\implies\\ ...

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@Xin Wang's last comment: In the first case you are simply, formally, looking at collection of k_max different, uncoupled oscillators. But you're only doing anything with the k'th one. k is an index in this case, nothing else but giving this specific oscillator a name. In the second case you only have one oscillator in your notation, so actually you don't ...

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First of all $\vec{J}$ is not an Hermitian operator from a Hilbert space to the same Hilbert space, its three components separately are. Therefore nothing requires that there must exist an orthonormal basis of eigenvectors of $\vec{J}$, that is a orthonormal basis of simultaneous eigenvectors of $J_x$, $J_y$, $J_z$. Otherwise these operators would commute ...

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Usually in many body theory these operators create and annihilate particles. There are different annihilation and creation operators for fermions and bosons (they obey different commutation relations). The states they act upon and the states "created" by them respect the required symmetries (antisymmetric for fermions, symmetric for bosons). The operators ...

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So I actually was able to figure this out! It turns out I skipped a step. So you begin with: $\langle\tilde{x}'\mid X \mid \tilde{x}\rangle = e^{-ig(x')/\hbar}\langle x'\mid X \mid x\rangle e^{ig(x)/\hbar}$ = $\langle x'\mid X \mid x\rangle e^{i(g(x)-g(x'))/\hbar}$ Next since $X\mid x\rangle = x\mid x\rangle$, the previous formula becomes $x\langle x'\mid ... 2 Firstly, I'd like to recommend Conformal Field Theory by Di-Francesco, it is a comprehensive text which is thorough and contains many applications of conformal field theory. The text is indispensable. In conformal field theory, it is often characteristic of correlation functions to diverge as points of two or more fields coincide. The operator product ... 2$R(\phi_0,k)$is the operator that rotates your co-ordinate system. But it is not suitable to apply a 2x2 matrix, as in this case, to a vector in Hilbert space. Mind that Hilbert space is unlike an ordinary orthogonal position space. Thus, using$U$, you map the operator into an equivalent operator which can operate on vectors in Hilbert space. 2 You're asking what could be a very deep question. But let me take a not-so-deep approach to answering that gets at the nature of science. It is experimental fact that we only measure particular values of quantities at the quantum level. That's it. Nature doesn't care about eigenvalues or operators; only humans do. That being said, physicists have, based on ... 1 Mathematics is a very powerful discipline. One can have a very large number of theories, starting from algebra, going to topological theories, and they are based on an internal very strict self consistency. They start by axioms, develop functional dependences and the whole mathematical construct is self contained . That is why one has proofs in mathematical ... 1 So let's remember that$X$is an operator, and$\langle X \rangle$is just a number, and we can use the definition of the expectation value$\langle O \rangle = \langle \psi | O | \psi \rangle$to work this out. \begin{eqnarray}\Delta X^2 =& \langle X^2 - 2X\langle X \rangle + \langle X \rangle ^2 \rangle \\ =& \langle X^2 \rangle -\langle ... 1 Your claim that the derivative is in the expansion of$\langle x'|\hat{p}\hat{x}|\psi\rangle$acts on everythin on the right is correct. Realize that $$\langle x|\hat{p}|\psi\rangle=\frac{\hbar}{i}\frac{d}{dx}\langle x|\psi\rangle=\frac{\hbar}{i}\frac{d}{dx}\psi(x)$$ So $$\langle x'|\hat{p}\hat{x}|\psi\rangle=\frac{\hbar}{i}\frac{d}{dx'}\langle ... 1 Actually, you got the right result, you just didn't see it. In the sense of distributions we have that$$x\delta(x-x')e^{i(g(x) - g(x'))/\hbar}=x\delta(x-x')$$These expressions, make no sense if not under an integration sign with a test function h\in C^\infty_0(\Bbb{R}) or h\in\mathcal{S}(\Bbb{R}), the smooth (infinitely differentiable) functions of ... 1 The Hilbert space of the states in this case is the Fock space. It is a linear space "constructed" by acting by the creation operators a^\dagger_k on the vacuum state |0\rangle, which has the property a_k|0\rangle=0. All other states are related so that the commutation relations between a,a^\dagger are satisfied. The individual states like ... 1 Let's take the spin, it's the simplest case, Q = \mathbf{s}. The operator \mathbf{s} is a vector, $$s = \mathbf{i}s_x + \mathbf{j}s_y + \mathbf{k}s_z$$ while the operator s^2 is a scalar $$s^2 = s_x^2 + s_y^2 + s_z^2$$. The operators s_x, s_y, and s_z don't commute two by two, but s_x^2, ... 1 In the first part of the linked document are not giving a formal derivation of the uncertainty principle. It is giving a particular example to show the general idea. The gaussian wavefunction is chosen because it is a particularly simple and happens to exactly satisfy the lower bound \sigma_x\sigma_p = \frac{1}{2}\hbar. For the harmonic oscillator being ... 1 I think you either misunderstood your lecturer or he used a very bad and uncommon notation. I would claim the expression a_k^{\dagger} |1,...,N \rangle = \sqrt{N+1} |k,1,...,N \rangle is simply wrong when understood within the usual notations. It is unclear what the state |1,...,N \rangle denotes. Now my best guess is that it is in 'first ... 1 Here are the two most important tricks: Commutators via derivative If A is an operator which has been written as a normal ordered^{[1]} product of a and a^\dagger, then the following is true$$[a, A] = \frac{\partial A}{\partial a^\dagger} \quad \textrm{and} \quad [a^\dagger, A] = -\frac{\partial A}{\partial a} \, .$\$ Wick's theorem (boson ...

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