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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 \begin{equation}|P\rangle = \sum ...


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 ...


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 ...


4

The result can be proved in a general way using, well, math. In particular the theory of semigroups of linear operators on Banach spaces (I know that seems advanced and maybe not physical, but it is an elegant way of proving the result you seek ;-) ). Define the Banach space $\mathscr{L}^1(\mathscr{H})$ of trace class operators over a separable Hilbert ...


4

CAUTION - ANSWER INCOMPLETE There is a gap in my argument (see the send); it relies on the claim that \begin{align} - \hat O^\dagger \hat A = \hat A\hat O \end{align} for all hermitian $\hat A$ implies $\hat O = 0$ which may not be true. Please comment if you know how to prove this or know of a counterexample. Update. Actually the claim above is ...


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}]\, ...


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 ...


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 \begin{equation} \left\langle \delta_{\varepsilon\, a}\right\rangle = 1\Pr(|x\rangle = |a\rangle) + 0\Pr(|x\rangle \perp |a\rangle) = ...


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, ...


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 ...


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 ...


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 ...


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. ...


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 ...


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


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 ...


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 ...


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 ...


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} ...


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 ...



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