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2

You have to interpret $|\frac{d}{dq} \psi\rangle$. Knowing that decomposition of the basis $|q'\rangle$ gives : $$|\psi\rangle = \int dq' \psi(q') |q'\rangle \tag{1}$$ You have : $$|\frac{d}{dq}\psi\rangle = \int dq' \frac {d\psi(q')}{dq'} |q'\rangle\tag{2}$$ So, applying it to $|\psi\rangle = |q"\rangle = \int dq' \delta(q"-q') |q'\rangle$, you get : ...

2

The operator $d/dx$ isn't a "function" and Dirac surely never claims so. It's an operator, something that changes one function to another. By a function, we mean something that maps one number to another. Functions of $x$, like $f(x)$, may also be connected with operators on the space of (wave) functions. The wave function $\psi(x)$ is mapped to ...

3

(Disclaimer: The more rigourously inclined individual may be better suited by looking at the Stone-von Neumann theorem, as Qmechanic notes) One can deduce that the momentum operator takes the form $\hat p = -\mathrm{i}\hbar\partial_x$ in the position representation from the fact that the momentum operator generates the infinitesimal translations as ...

2

I) Comment to the question (v1): The Schrödinger position representation $$\hat{p}_k ~=~ \frac{\hbar}{i} \frac{\partial }{\partial x^k}, \qquad \hat{x}^j ~=~x^j,$$ correctly reproduces the canonical commutation relations $$[\hat{x}^j,\hat{p}_k ]~=~i\hbar ~\delta^j_k ~{\bf 1},$$ while the proposal $$\hat{p}_k ~=~ \frac{\hbar}{i} \frac{1}{x^k}, ... 1 You can find derivation of these operators in most standard quantum mechanics textbooks. For your convenience, see https://en.wikipedia.org/wiki/Momentum_operator and https://en.wikipedia.org/wiki/Position_operator. For the second question, Paul Dirac said in his classic The Principles of Quantum Mechanics: A measurement always causes the system to jump ... 1 The quantum field has nothing to do with the wavefunction. This is a peculiar confusion that seems to arise quite often. The wavefunction \psi(x) is a way of representing a quantum state \lvert \psi \rangle in a Hilbert space \mathcal{H}_{\mathrm{QM}} that is equipped with a position basis \{\lvert x \rangle \rvert x \in \mathbb{R}\} by setting ... 1 The answer to both questions is that D act on Hilbert space states. I'll answer them in reverse order. what exactly do we mean by the eigenvectors of D? Are they fields in space-time? No, in this context, eigenvectors of D are states living in the Hilbert space of the field theory. Because it is only in this sense that the commutation relations between ... 2 Q is not a functional, but a linear operator. Since it is linear, there are no problems in using the chain rule. I will pretend there are no domain problems here, and that you can exchange limits and integrals as you wish (e.g. I'm supposing you can use the dominated convergence theorem), and obviously that each quantity is differentiable. ... 4 The commutation relations$$ [D,P_{\mu}] = +i P_{\mu} , \qquad [D,K_{\mu}] = -i K_{\mu} $$show that P_{\mu} and K_{\mu} raise and lower the conformal dimension of a state. In other words, if you have a state |\phi\rangle of dimensions \Delta, so that D\, |\phi\rangle = i\Delta |\phi\rangle, then$$ D \, P_{\mu} \, |\phi\rangle = [D,P_{\mu}]\, ...

2

$$<\hat{A}> = \int \psi^*(x)\hat{A}\psi(x) dx$$ now $\hat{A}\psi(x)=a(x)\psi(x)$ so, $$<\hat{A}>=\int a(x)|\psi(x)|^2dx$$ Let $|\psi(x)|^2dx = d \mu$ now $\int|\psi(x)|^2dx=\int d\mu = 1$ $$<\hat{A}> = \int a(\mu)d\mu$$

0

Let us use bra-ket notation. Suppose that the operator $\hat A$ has discrete and bounded eigenvalues $\mathcal A =\{A_1, A_2, \ldots\}$ with eigenkets $|A_1\rangle, A_2\rangle,\ldots$. The eigenkets form a complete orthogonal set since $\hat A$ is symmetric. Then any ket $|\psi\rangle$ can be expanded, $$|\psi\rangle = \sum \psi_n |A_n\rangle.$$ Since the ...

0

Observables are associated with linear operators, not measureable functions, so how can we talk about the expectation of a linear operator? The "expectation of a linear operator" is a term from quantum theory. It is defined by the integral $$\int \psi^*(x)\hat{A}\psi(x) dx$$ or similar. The meaning of this number is not necessarily the same as ...

1

There are two answers to this. One answer simply points out that the probability of the jth outcome specified by the Born rule $p_j = tr(\rho\hat{P}_j)$, where $\hat{P}_j$ is the projector onto the jth outcome, satisfy the axioms of probability: http://mathworld.wolfram.com/ProbabilityAxioms.html. Another answer is that the Born rule can be explained ...

2

Since you want a bit of mathematical rigor: A quantum state is a self-adjoint positive trace class operator on a Hilbert space with trace 1. This is called density matrix $\rho$. In its simplest form, given $\psi\in \mathscr{H}$, $\rho$ is the orthogonal projector on the subspace spanned by $\psi$. Let $E_\rho(\cdot):D_\rho\subset\mathcal{A}(\mathscr{H})\to ... 1 You are maybe making confusion between the action of an observable (operator), and the measurement process. In particular:$A\psi$is simply a vector of the Hilbert space. In my opinion it has not much sense of talking about "initial" and "terminal" state because you are not looking at a dynamical situation. If you want to know the average value of an ... 1 From the equations,$\phi$is the operator acting on the variable/state$\xi$. It is important to notice also that in the factorization the$a_i$numbers are real and the$c_i$are complex. This factorization comes just from the mathematical fact that for a polynomial equation of degree$n$, such as the first equation, there are$n$complex roots. 1 No$\hat\phi|0\rangle$is not an eigenvector of$\hat\phi$. You can see this, for example, by writing out$\hat\phi$in terms of creation and annihilation operators, then compare$\hat\phi|0\rangle$against$\hat\phi^2|0\rangle$, and observe that one is not a scalar multiple of the other. So as you suspected, eq. 5 is not correct To obtain some analogy of ... 3 You can utilize the construction of the$Q$space, as described in Reed and Simon vol.2, page 228-230. Oversimplifying, you can make the analogy$\lvert \phi\rangle \sim \lvert x\rangle$, but the associated momentum is not$\hat{p}$, but$\hat{\pi}$(the canonical conjugate momentum of the field$\hat{\phi}$). With slightly more precision: the Fock space ... 3 $$D(\hat{X}_i):= \left\{ \psi \in L^2(K, d^nx)\:\left|\: \int_K x_i^2|\psi(x)|^2 d^nx< \right.+\infty \right\} = L^2(K, d^nx)$$ where the last identity holds true if$K$is bounded (in particular compact) because$x_i^2$is bounded as well thereon (in this case the operator is bounded, too). Moreover $$D(\hat{J}^2):= \left\{\psi \in L^2(\mathbb S^2, ... 0 To answer the question I think you meant to ask, V(x,t) is a real-valued function. But you can also think of it as an operator. An operator is anything that maps functions to other functions, and multiplication by a fixed function is one way to do that. To be a little more pedantic with the notation, if you use \hat{V} to represent the operator, then ... 0 As usually, physicists are being a bit sloopy with notation here. Mathematically, there's a big difference between a function (such as V) and a function's value for some arguments (such as V(x,t)). The latter is indeed simply a number, scalar if you like. However, in physics, functions are often used just as "families of values" and values with some free ... 3 In the representation given by you, the V(x,t) is a scalar function depending on x and t. However, you already have choses a basis (the x-basis). In general, the \hat{V} or rather \hat{H} in the Schödinger Equation is an operator. This operator can then be evaluated in a basis of your choice. If you are familliar to the dirac notation one can ... 3 The potential is, without any ambiguity, an operator. There is no other way, because it changes the spatial dependence of the wavefunction:$$ \Psi(x,t)\text{ changes to }V(x,t)\Psi(x,t). $$It's unclear to me what you mean by "scalar", though, because this word can have multiple meanings. For example, it is indeed a scalar operator - as opposed to vector ... 0 Answer to question one : The Principle of Quantum Mechanics by R. Shankar page 149 reads "Barring a few exceptions, the schrodinger equation is always solved in a particular basis. Although all basis are equal mathematically, some are more equal that others. First of all, since H = H(X,P) the X and P basis recommend themselves....The choice between the two ... 0 You have to clarify the term "basis" in infinite dimensions. By definition, finite-dimensional spaces have a finite basis (so that, for instance, \mathbb{R}^n\approx\mathbb{R}^m iff n=m). But for infinite-dimensional spaces, there are various types of bases (assuming you can take infinite sums). A natural one (using finite sums) is the Hamel basis, but ... 5 First, the term "basis spaces" isn't standard in quantum physics but let us assume that we understand what the sentences approximately mean. Second, the momentum is continuous (not quantized) if the position space is noncompact (infinite). The momentum only becomes quantized if the position space is compact (or periodic), and indeed, it's been ... 2 An observable is a self-adjoint operator \mathcal{O} on the Hilbert space of states \mathcal{H}. The spectral theorem tells us that such an operator has an orthonormal basis of eigenvectors in \mathcal{H}, if it is compact. If it is not compact, we have to "enlarge" the Hilbert space to something called rigged Hilbert space or Gelfand tripel. A ... 0 Unfortunately not every operator, which is used to get an observable gives you enough elements to make a base of the Hilbert space. As far as I know every eigenstate is orthogonal to each other. But the “length” could be arbitrary. Mostly used eigenstates in quantum mechanics are normalized for better usage. 3 The Projection Operator \delta_{\varepsilon\, a} is 1 for some particular |a\rangle and is 0 on any orthogonal state. The definition of the expectation in classical probability theory for this operator is given by \left\langle \delta_{\varepsilon\, a}\right\rangle = 1\Pr(|x\rangle = |a\rangle) + 0\Pr(|x\rangle \perp |a\rangle) = ... 4 The position operator in three dimensions is a vector operator, which, as in mpv's answer, acts as$$ \hat{\mathbf r}\psi(\mathbf r)=\mathbf r \psi(\mathbf r). $$Note that the hat denotes an operator and not a unit vector. The definition of a vector operator is somewhat tricky, and it is indeed startling that an operator can have vector eigenvalues. The ... 3 The position operator in 3D is a vector in 3D:$$ \hat{\bf r} \psi({\bf r}) = {\bf r} \psi({\bf r}) $$See here. 2 The first equation is quite elementary to derive. First, we define the expectation value of an operator:$$\langle O \rangle=\langle \psi(t)| O | \psi (t)\rangle = \langle \psi(t=0)|U^\dagger O \ U |\psi (t=0)\rangle \tag{1}$$where U is the time evolution operator: U=\exp\left( -i\frac{Ht}{\hbar}\right) if H is independent of time. Now, we can take ... 2 The second one is a particular case of the first one, as I understand it. In the first one we are asuming that, in general, Q can depend on time explicitily and through other dynamical variables, as for example, position or momentum. That is,$$ \hat{Q} = \hat{Q}(\hat{x}(t),\hat{p}(t),t) $$So, if we want to know how it evolves with time, we must to know ... 1 For your second problem, the propagator can be written with its indices as$$ (S_F)_{\alpha\beta}(x-y) = \langle T\{\psi_{\alpha}(x)\bar\psi_{\beta}(y)\} \rangle $$Then we have$$ \langle T\{\bar\psi(x)\Gamma\psi(y)\} \rangle = \Gamma_{\alpha\beta} \langle T\{\bar\psi_{\alpha}(x)\psi_{\beta}(y)\} \rangle = -\Gamma_{\alpha\beta} ... 6 Dirac being opaque and hard to follow? Well I never... In Chapter 10 Dirac argues on physical grounds that the eigenkets of an observable must form a complete set. His argument goes that say we have an observable with eigenstates$|\varepsilon\rangle$and some general state$|P\rangle$. Then I can write$|P\rangle$as |P\rangle = \sum ... 0 I) Before we start let us briefly recall certain aspects of the formalism from Ref. 1. The Minkowski sign convention is$(+,-,-,-)$. The 'phase space' measure for a particle$A$is $$d\tilde{k}~:=~ \frac{d^3k}{(2\pi)^3 2\omega_{k,A}} ~=~ \frac{d^4k}{(2\pi)^3} \delta(k^2-m^2_A)\theta(k^0),$$ $$\omega_{k,A}~:=~\sqrt{{\bf k}^2+m^2_A}~>0~. \tag{3-35}$$ ... 0 You obtain this by Wick's Theorem, which can be stated as $$T\{\phi_1\phi_2...\phi_n\}=N\{\phi_1\phi_2...\phi_n+\sum\text{all possible contractions of }\phi_1\phi_2...\phi_n\}$$ where N is the normal ordering operator which puts all the daggered fields on the left ( for example ... 1 Recall the Heisenberg principle which tells us that a we cannot know both position and momentum at the same time $$\Delta x \Delta p \geq \hbar/2$$ For a particle, we have actually at most$\Delta x = L$, so that we cannot in principle have a sharp momentum state. Fortunately, there are eigenstates which are eigenstates of$\hat{P}^2$, that is, states ... 1 Taking the standard$[0,L]$problem, eigenfunction and energy eigenvalues are: $$\varphi_n=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}, \qquad E_n=\frac{\hbar^2\pi^2n^2}{2mL^2}.$$ This means that stationary 1D box systems (e.g. insulated ones) only admit states with a discrete set of possible energies, as above. Now, that as far energy is concerned. What about ... 0 Your lecture notes, or your transcription of them, are in error. You should have $$\hat{L_{+}}|\lambda,m_\mathrm{max}\rangle= 0 \\ \hat{L_{-}}|\lambda,m_\mathrm{min}\rangle= 0$$ That is, raising the maximum-projected state doesn't give you$\left|\lambda,m\right> = \left|0,0\right>$, a state with no angular momentum, and it doesn't give you a vacuum ... 0 Sometimes it's helpful to consider finite dimensional special case to get insights on a problems like these. Let$P$and$L$be in$\mathbb{C}^{n \times n}$. If$P^2 = P$, then$Pv = \lambda v$leads to$\lambda^2 = \lambda$, i.e., the eigenvalues are in$\{0, 1\}\$. In the context of quantum, we consider Hermitian operators. In this case, we can find some ...

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