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In general, $\frac{A}{B}$ is lazy physicist shorthand notation for $B^{-1}A$. You might rightly complain that there is an ordering ambiguity and the expression could also mean $A B^{-1}$. That's completely correct, and this notation is only meaningful if $[A,B^{-1}]=0$. If we take $A=g_o a^\dagger a$ and $B = w_m - g_{cK} a^\dagger a$, then indeed $[A, B^{-1}... 11 In general, you cannot rewrite$\langle a | M | b \rangle$as$|b\rangle \langle a| M$. You can see that the two are not the same by just comparing what type of mathematical entity they are:$\langle a | M | b \rangle$is a matrix element (of the operator$M$), which is a (complex) number. On the other hand,$|b \rangle \langle a|$is an operator, as is$M$, ... 9 The expressions you write are extensions in infinite-dimensional Hilbert space of plain matrix expressions. Their analogs for finite dimensional real vector spaces and their matrices indexed by a finite set of indices i,j, whose repeated form implies summation over the whole set, are $$|a\rangle ~~\mapsto ~~ a_i \\ |b\rangle ~~\mapsto ~~b_i\\ M~~\mapsto M_{... 7 If$$ \sum_k |\psi_k\rangle \langle \psi_k|= {\mathbb I} $$then, sandwiching this expression between \langle x| and |x'\rangle, we have$$ \sum_k \langle x|\psi_k\rangle \langle \psi_k|x'\rangle= \langle x|x'\rangle $$or, since \langle x|\psi_k\rangle\equiv \psi_k(x) and \langle \psi_k|x'\rangle = (\langle x|\psi_k\rangle)^*, and also \langle x|... 7 when I operate it on a state |ψ⟩ there will be some error every time I measure it This is a common misconception. Performing a measurement on a system is not mathematically represented by applying the corresponding Hermitian operator to that state. The operator just tells us the possible measurement outcomes through its eigenvalues, and it allows us to ... 5 People call it an operator. It is a thing which would map a function to another function. It's a lot like how a matrix maps a vector to another vector. You can invent lots of operators, like$$2$$or$$2 + 2x$$or$$2 + 2x +\frac{d}{dx}$$which, when acting on a function f(x), would give$$2 f(x) + 2x f(x) + \frac{df}{dx}$$by definition. There are some ... 5 In general, we can say that C=AB will have real, imaginary and complex eigenvalues (complex of the form z=a+ib where and \{a,b\in \mathbb{R}\mid a,b \ne 0\} as shown in the comments by Mark and Qmechanic's answer). For example, if$$A=\begin{bmatrix} 0 &1 \\ 1& 0 \end{bmatrix}\ \ \text{and}\ \ B=\begin{bmatrix} 1 & 0\\ 0& -1 \end{... 4 To supplement Andrew's nice answer, note that vectors in the occupation number basis are eigenvectors of$a^\dagger a$, i.e. $$a^\dagger a |n\rangle = n|n\rangle$$ As a result, the action of your operator$\hat \xi$on such a vector becomes $$\hat \xi|n\rangle = \frac{g_0 n}{w_m - g_c n}|n\rangle$$ which extends by linearity to any state$|\psi\rangle = \...

4

It turns out that the cyclic property of the trace, i.e., $$\operatorname{Tr}(ABC) = \operatorname{Tr}(CAB) = \operatorname{Tr}(BCA),$$ holds when the matrices in question are not square. In particular, interpreting the kets and bras as row and column vectors, respectively, one can write \operatorname{Tr}(|a\rangle\langle a|X|b\rangle\langle b|) = \... 4 All you need is {\rm tr}\{X\}=\sum_a \langle a|X|a\rangle, where |a\rangle is a compete set of orthonormal states. 4 Going off the quote of Lanczos found by Conifold... "In spite of the radical departure of [quantum] concepts from those of the older physics, the basic feature of the differential equations of wave-mechanics is their self-adjoint character, which means that they are derivable from a variational principle. Hence, in spite of all differences in the ... 4 We have \begin{align} j(z) : e^{ i k X } : &= \frac{i}{\alpha'} \partial X(z) \sum_{n=0}^n \frac{ : ( i k X )^n : }{ n! } \\ &= \frac{i}{\alpha'} \sum_{n=0}^n \frac{ ( i k)^n }{ n! } \partial X(z) : X^n : \\ &\sim \frac{i}{\alpha'} \sum_{n=1}^n \frac{ ( i k)^n }{ n! } n : X^{n-1} : \partial_z X(z) X(0,0) \\ &\sim \frac{i}{\alpha'} \... 4 Is there a geometrical interpretation for this? Yes. Without being too rigorous here, if you identify an operator with a matrix, and a ket with a column vector, then from linear algebra, when you multiply a matrix with a basis column vector (assuming the standard basis), it just picks out the column in the matrix corresponding to the index of the non-zero ... 4 Note: I use Einstein summation notation throughout. Any expression which has an index repeated twice should be understood to have an omitted, but implicit summation over that index. Setup Consider the expression T|v\rangle $$where |v\rangle \in \mathcal{H} is an abstract, finite-dimensional vector living in a Hilbert space and T : \mathcal{H} \... 4 TL;DR: Assuming that A,B are self-adjoint, the product AB does not need to be diagonalizable. And if AB is diagonalizable, the eigenvalues need not be real or imaginary. Example 1: AB is not diagonalizable:$$A~=~\begin{pmatrix} 0 & 1 \cr 1 & 0 \end{pmatrix} \quad\wedge\quad B~=~\begin{pmatrix} 1 & 0 \cr 0 & 0 \end{pmatrix}\quad\...

4

Suppose $A$ and $B$ are Hermitian operators: what is the nature of the eigenvalues of $[A,B]+\{B,A\}=2AB$? First, "$A$ is Hermitian" from the practical point of view means that: $A=A^\dagger$, it is unitarily diagonalizable and has real eigenvalues. The same for $B$. Now, $(AB)^\dagger = B^\dagger A^\dagger =BA$: this tells us that, in general $AB\... 4 In general $$\langle \varphi| A |\psi \rangle^* = \langle \psi| A^\dagger |\varphi \rangle$$ If$A$is hermitian$\Rightarrow A = A^\dagger\begin{align} &\Rightarrow \langle \varphi| A |\psi \rangle^* = \langle \psi| A |\varphi \rangle\\ &\iff (|\psi \rangle)^\dagger A^\dagger (\langle \varphi|)^\dagger = \langle \psi| A ... 4 There is always uncertainty associated with measurements,a nd which is analyzed using statistical methods. This uncertainty however can be made (in principle) vanishingly small with improving equipment. I say in principle, because there are obviously some limitations - e.g., our measurement cannot last longer than the lifetime of the universe, etc. The ... 3 A less subtle alternative to using the cyclic property is "brute forcing" it, by plugging in the definition of the trace. It goes like this, \begin{aligned} {\rm Tr}(\left |a\rangle \langle a|X|b\rangle \langle b \right|) &= \sum_n\langle n | \big(|a\rangle \langle a|X|b\rangle \langle b| \big) |n\rangle \\ &= \sum_n\langle n |a\rangle \... 3 As you may know, the metric tensor is a bilinear 2-form. It accepts two vectors from vector space V and gives back a real number in \mathbb R. It is linear in both arguments, hence 'bilinear'. The metric tensor is interpreted as a linear operator in the sense that it maps one of its arguments (either one; doesn't matter because it's symmetric) to a dual ... 3 The metric tensor is bilinear by definition, since tensors are multilinear maps from vector spaces (and their dual spaces) to real numbers, which means that they must be linear in all slots. In general, on a n-dimensional pseudo-Riemannian manifold, the metric tensor cannot be put into diagonal form everywhere by a coordinate basis. It can only be done if ... 3 One can always think of an operator as a matrix and refer to the corresponding matrix methods. Since the operator in question is non-hermitian, its right and left eigenvectors are generally not the same, and the diagonalizing transformation is generally not a unitary one, i.e., one cannot restrict it to matrices satisfying S^\dagger=S. The more general ... 3 I am ignoring \frac{\partial}{\partial t}, since it does not involve the issue that confuses you. You are right to be puzzled by the abuse of notation, \hat p \sim -i\hbar \frac{d}{dx}, but your teacher should have made this very clear right from the start. What it means is that this is a representation in the coordinate picture, which is to say \hat ... 3 As @Cream points out in a comment, no, you do not have enough information to draw that conclusion. What you are asking for is if: $$[[\hat{A},\hat{B}], \hat{A}] = 0 \quad \quad \stackrel{\color{red}{?}}{\implies} \quad \quad [\hat{A}, \hat{B}] = 0.$$ But what you have "shown" instead is that: $$[\hat{A},\hat{B}] = 0 \quad \quad \implies \quad \quad ... 3 Never forget to think about statistics when considering quantum mechanics. Your question is related to the three correlations between pairs of three variables. Famously, "is correlated with" isn't as transitive as we'd like to think. Let |\chi\rangle:=\sum_m|m\rangle. Write \sim between complex numbers of the same modulus. We can choose three ... 3 Simply note that [a,a^\dagger]=1 \implies aa^\dagger = 1+a^\dagger a. 3 We'd need additional context but one interpretation is that you get a second eigenvalue to distinguish between those eigenstates with the same eigenvalue of A, i.e the states with eigenvalue \alpha go from arbitrary labelling$$ \{\vert \alpha,i\rangle, i=1,\ldots, n\}\tag{1} $$to$$ \{\vert \alpha,\beta_1\rangle, \vert \alpha,\beta_2\rangle\ldots, \... 3 I will approach your problem from a pure mathematical perspective: Let me first say that in your post you use the same notation\dagger\$ for two distinct mappings which have two different definitions and notations. One mapping acts on the space of linear densely defined operators in a complex separable Hilbert space and the other mapping is from the ...

3

In classical QM, $$\partial_x = i \hbar^{-1} p.$$ However, the time derivative is (According to the Schrodinger equation) $$\partial_t = -i \hbar^{-1} H.$$ Whilst spatial translations are generated by the momentum, time shifts are actually generated by minus the energy (or depending on your definitions, spatial translations are generated by minus the ...

3

No, you should not think of the uncertainty principle as meaning that measurements necessarily have associated errors- it is more fundamental than that. What the uncertainty principle means is that there are certain combinations of properties- such as position and momentum- for which a particle can never have simultaneous exact values. It is not that the ...

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