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2

Without loss of generality, let's take the $|\lambda_i\rangle$ to be orthonormal. Notice that, by the spectral theorem, the hamiltonian can be written as follows: $$H = \sum_i \lambda_i P_i, \qquad P_i = |\lambda_i\rangle\langle \lambda_i|$$ Each operator $P_i$ is a projectors onto the subspace spanned by $|\lambda_i\rangle$. Notice, in particular, ...

7

Starting with: $$U(t,t_i) = e^{\frac{-i}{\hbar }H(t-t_i)}$$ If $t_i=0$: $$U(t,0) = e^{\frac{-i}{\hbar }Ht}$$ Using the identity: $\sum\limits_i \left|\lambda_i\right>\left<\lambda_i\right|=\mathbb{I}$ $$U(t,0) = \sum\limits_i e^{\frac{-i}{\hbar }Ht}\left|\lambda_i\right>\left<\lambda_i\right|$$ Since the exponential of an operator is (by ...

1

You can prove this by induction. I'll drop the operator hats as they're a pain to write. First step: Suppose $\hat H = \hat q_k$. Then $[ H, p_i] = [q_k, p_i] = i\hbar \delta_{ik} = -\frac{\hbar}{i} \frac{\partial H}{\partial q_i}$ So in the special case that the Hamiltonian has this form, the claim is true! Now suppose that the claim is true for $H = ... 2 If you are acquainted to matrices, then$|\psi\rangle$is very much like column vector, and$\langle\psi|$is similar to row vector. Operators correspond to square matrices. The conjugate transpose$^\dagger$is similar to the matrix transpose$^\mathrm{T}$in the sense that it turns columns to rows and vice versa. Then (neither of your 1-3 lines), if we ... 3 You're not getting your facts right at all. How do we know from this$\langle W \rangle = \int_{-\infty}^{\infty} \bar{\Psi}\left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + W_p \right) \Psi dx$or this$\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + W_p$that we have an eigenfunctiuion and eigenvalue. Answer: we don't. All I know about operator ... 4 First i need some explaination on how do we know this? It's stipulated. Maybe it will help your understanding if we phrase it this way: Let$\psi$be an eigenfunction of an operator$\hat{H}$with eigenvalue$W$. (Update to address OPs comment). Spectral Theorem: Theorem. There exists an orthonormal basis of V consisting of eigenvectors ... 0 This is only true if A and B are independent. In that case,$\mathbb{E}[AB] = \mathbb{E}[A]\mathbb{E}[B]$and your calculations are correct. Without knowing the distribution for B, this is the strongest supposition needed for your calculations to be correct. 3 The commutators in the above expressions are sued to change the order of the Hamiltonian and annihilation or creation operators. I'll show you the first one in some detail, the second one should not give you problems afterwards. We start from$\hat{H}\hat{a}\psi_n$. Using the commutator$[\hat{H},\hat{a}] = \hat{H}\hat{a}-\hat{a}\hat{H} = ...

2

Let us change OP's notation $a\to a_1$ and $b \to a_2$. We write collectively the two annihilation operators as a column two-vector $$\tag{1} \vec{a}~:=~\begin{bmatrix} a_1 \\ a_2 \end{bmatrix}.$$ We have the Heisenberg algebra $$\tag{2} [a_i,a_j^{\dagger}] ~=~\delta_{ij} {\bf 1}\qquad i,j~\in~\{1,2\}.$$ and the vacuum state $$\tag{3} a_i | ... 3 There are many ways to go around this. You can start from the coherent states and apply the unitary \hat{O} directly on them. That will not be that simple because you will get a term \hat{O}e^{\alpha\hat{a}^\dagger}e^{\beta\hat{b}^\dagger}. Now, the typical approach would be to exchange the order of the operators to get something like ... 3 Start with your \hat{H} = \hbar \omega \left( \hat{a}^\dagger\hat{a} + \frac{1}{2} \right). I will omit hat notation from this point. The commutator then reads as \left[ H, a \right] = \hbar \omega \left[ \left( \hat{a}^\dagger\hat{a} + \frac{1}{2} \right) a - a \left( \hat{a}^\dagger\hat{a} + \frac{1}{2} \right) \right] = \hbar \omega ... 3 On the Wikipedia page you link to there is a derivation of the commutation relation between \hat{a} and \hat{a}^{\dagger},$$ [\hat{a},\hat{a}^{\dagger}] = 1.$$This directly leads to (use the relation [AB,C]=[A,C]B+A[B,C])$$[\hat{a}^{\dagger}\hat{a},\hat{a}] = -\hat{a} , \qquad [\hat{a}^{\dagger}\hat{a},\hat{a}^{\dagger}] = +\hat{a}^{\dagger}.$$Up ... 3 I) OP's factor \frac{1}{2} comes from using the truncated version$$\tag{1} e^Ae^B~=~e^{A+B+\frac{1}{2}[A,B]}$$of the Baker-Campbell-Hausdorff formula. Formula (1) holds if the commutator [A,B] commutes with both the operators A and B. II) What is actually needed in OP's calculation is rather this version$$\tag{2} e^Ae^B~=~e^{[A,B]}e^Be^A$$... 2 If you add the hypothesis of unitarity you have that the conformal weights of your field \Delta_i must obey the following inequality:$$\Delta_i \geq \frac{d-2}{2} + l$$where l is the spin of the operator. The conformal bootstrap equation give you also constraints on the coefficients C_{ijk} appearing in the 3-point correlation function. As far as I ... 1 The derivation by Sakurai is by no means mathematicaly rigorous, so you should expect something like your argument about the scalar product. Indeed, we have everything more or less fine until$$ [x,\mathcal{T}(\epsilon)]|z\rangle=\epsilon|z+\epsilon\rangle $$where we want to replace |z+\epsilon\rangle by |z\rangle and claim that it is ok in the first ... 1 Here's the most logical way to proceed if you ask me. Given any a\in\mathbb R, we define the translation operator T_a by its action on position basis vectors$$ T_a|x\rangle = |x + a\rangle $$One can prove the following properties: T_a is unitary for each a\in\mathbb R. T_aT_b = T_{a+b} for all a,b\in\mathbb R. It follows (by Stone's ... 4 Spectral geometry is one of the many ways mathematicians think about geometry. The general idea is that if you have some manifold equipped with a metric, you can cook up some canonical differential operators. These operators can be thought of as linear operators, acting on (infinite-dimensional) vector spaces of functions, tensors, spinors, and the like. ... 1 Main point: You should allow the possibility of sign factors appearing into the definition of the Hilbert space representation of fermionic operators, cf. fermionic Fock space. In more detail, consider the CAR algebra$$\tag{1} \{c_{\sigma}, c_{\tau}\}~=~0, \qquad \{c_{\sigma}, c^{\dagger}_{\tau}\}~=~\hbar {\bf 1}, \qquad\{c^{\dagger}_{\sigma}, ...

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I think you are right. Using really simple commutator math. All you need is this: $$[AB,C] = A[B,C] + [A,C]B$$ Then in your case: $$A=P$$ $$B=XP$$ $$C=P$$ $$[PXP,P] = PX [P,P] + [P,P]XP = PX[P,P] + P[X,P]P + [P,P] XP$$ As you said, [P,P] is antisymmetric to itself, and therefore we can remove all the [p,p] terms. We then have left only one term: ...

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You teacher seems to have made a mistake. I imagine that he/she did something like this: \begin{align} [PXP, P] &= P[XP,P]+[PX,P]P \\ &= P(X[P,P]+[X,P]P)+(P[X,P]+[P,P]X)P \\ &= P[X,P]P+P[X,P]P \\ &= 2i\hbar P^2 \end{align} Notice that the first equality is wrong. You can't peel operators off to the left and right if there are three ...

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Notice that \begin{align} i\frac{d(\psi^*\psi)}{dx} &=\frac{d\big[(-i\psi)^*\psi\big]}{dx} \\ &= \frac{d(-i\psi)^*}{dx}\psi + (-i\psi)^*\frac{d\psi}{dx} \\ &= \left(-i\frac{d\psi}{dx}\right)^*\psi + \psi^*\left(i\frac{d\psi}{dx}\right) \\ \end{align} Now subtract the second term on the right from both sides to get \begin{align} ...

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This is not as such related to field theory. The same story happens in regular quantum mechanics, but then just rotations or gallilei transformations. I do not think it is necessary to resort to Lie groups in order to understand what is going on. I would interpret $U(\Lambda)$ as the operator which translates a quantum state into the frame of reference ...

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The fact that symmetries are represented by unitary or anti-unitary operators is, indeed, due to Wigner's theorem. When we write $U(\Lambda)$ what we really mean is that $U$ is mapping from the Lorentz group to the set of unitary operators on the Hilbert space $\mathcal H$ of the quantum field theory; $$U: \mathrm{SO}(3,1)^+\to \mathrm U(\mathcal H)$$ ...

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Answering this question properly would require your understanding of the mathematics and principles behind Lie Algebras, group theory, and Field Theory. However, we are in the awkward predicament that knowing these things would automatically answer your question. Thus, you will have to forgive me in advance because I am not very good at summarizing in a few ...

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Brief explanation: When going from classical Lagrangian (say, non-relativistic point-)mechanics to quantum mechanics, there is an intermediate step known as classical Hamiltonian mechanics. To reach the intermediate step, one has to perform a Legendre transformation $(q,\dot{q}) \longrightarrow (q, p)$, where $(q, p)$ are (generalized) canonical phase ...

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