# Tag Info

7

Your problem is basically that of a charged particle in a magnetic field in the Landau gauge, but I'll explain the general theory for such systems as it's both interesting and suprisingly complicated --- and seldom found in textbooks. It's essentially the theory of Bose Bogoluibov transformations. Suppose we are given classical (or quantum) quadratic ...

6

This is the definition of a function of an observable. If $A$ is an observable and $A |a\rangle = a\, |a\rangle$, then a function $f$ of $A$ is defined through $$f(A) |a\rangle = f(a) |a\rangle \quad .$$ Thus, in the case of a single-particle potential $V(X)$, where $X$ denotes the position operator, we find $$V(X) |x\rangle = V(x)|x\rangle$$ and ...

5

Entanglement doesn't break locality. Entanglement must be generated locally. Wavefunction collapse a la Copenhagen interpretation breaks locality. QM and QFT are equivalent with respect to to locality. If you don't allow collapse in your interpretation then they are both unitary and local theory. If you do allow collapse then they are local until an ...

5

If the Hamiltonian is time-independent, then it seems much simpler to use $$\vert\Psi(t)\rangle = e^{-i \hat H t/\hbar}\vert\Psi(0)\rangle\, .$$ Then it is automatic that \begin{align} \langle \Psi(t)\vert\Psi(t)\rangle &= \langle \Psi(0)\vert e^{i\hat H t/\hbar} e^{-i \hat H t/\hbar}\vert \Psi(0)\rangle\, ,\\ &=\langle \Psi(0)\vert \Psi(0)\rangle ...

4

This turns out to be a surprisingly difficult experiment to do. To be absolutely sure you are seeing interference between different electrons you need two independent electron sources, and getting the two sources mutually coherent is very hard. As far as I know this was first achieved in 2007 as reported in Interference between two independent electrons: ...

4

Objects like your door do not "emit" any visible light. Instead they reflect the light that is coming from a lamp or the sun. If you put the door in total darkness it will be invisible to your eyes - unlike a piece of red-hot iron. Like all objects, the door does emit electromagnetic radiation as long as its temperature is above absolute zero. ...

4

Yes. Much like an entangled pair of photons should be treated as a single system whose wavefunction is a Bell state, the same ultimately applies to the universe. It is a system where everything is entangled and tracing out all degrees of freedom except maybe a handful of particles you're interested in is going to be an approximation. Whether it's a good or ...

3

Hints: The main point of the problem seems to be that with the given assumptions, i.e. that the wavefunction $\Psi\in C(\mathbb{R})$ is continuous, and that the potential $V$ is piecewise continuous with finite jumps/discontinuities, one can use a bootstrap argument to prove that $\Psi$ is in fact of class $C^1$ and piecewise twice differentiable, cf. e....

3

Let's start completely abstractly, and consider the abstract algebra over $\mathbb C$ generated by symbols $\sigma_1,\sigma_2,\sigma_3$ subject to your relations: $\{\sigma_i,\sigma_j\} = 2$, in particular $\sigma_i^2 = 1$. It is not hard to see that this algebra is generated as a vector space by the 8 elements $1, \sigma_1, \sigma_2, \sigma_3, \sigma_1\... 3 If the commutator is a negative number, we can always rescale the operators to have$[\alpha, \alpha^\dagger] = -1$. In that case, consider the operator$N = \alpha^\dagger \alphawhich is Hermitian. By the properties of the commutator \begin{align} [N, \alpha] &= [\alpha^\dagger, \alpha] \alpha = \alpha \\ [N, \alpha^\dagger] &= \alpha^\dagger [\... 3 There are several non-invasive medical imaging techniques that can detect activity in the brain at different levels of temporal and spatial resolution. These include magnetic resonance imaging, positron emission tomography, electroencephalography and magnetoencephalography. However, the electrical and magnetic fields produced by brain activity are very weak, ... 3 It does so by reflection. Light containing all wavelengths (colors) of light falls upon the door surface, and the atoms and molecules there absorb certain of those wavelengths and reflect others. The reflected colors are what your eyes perceive. 3 This definition of the adjoint operator is a direct application of Riesz' lemma. (I prefer another equivalent definition which does not use Riesz' lemma.) IfH$is a Hilbert space, a linear functional$H \ni x \to f(x)\in \mathbb{C}$is continuous if and only if it is of the form$f(x)=\langle x, z_f\rangle$for some$z_f\in H$and this element turns out to ... 2 This is more a question of how to normalize the eigenstates. This is just a convention that can choose. Myself, I like to write$\langle x|x'\rangle= \delta_g^n(x-x')where the delta function is defined by $$\int_M d^nx \sqrt{g} \,\delta^n_g(x-x')=1,$$ but other choices may be preferable. What is important that you state your normalization conventions ... 2 You evidently missed my point about trivial arithmetic checks on the Kronecker product reduction, in the convention where representations are labelled by their dimensionality, used here for SU(2) (in contrast to the atomic physics convention) and flavor SU(3) and GUTs in particle physics. In this convention, $$\mathbf{2}\otimes\mathbf{2}\otimes\mathbf{2}\... 2 For the \psi in the Dirac equation, interpreting the quantity |\psi(x)|^2 as a probability density doesn't work for other reasons, but the quantity |j^0(x)|^2\propto |\psi(x)|^2 is still meaningful as a charge density. The total charge should be Lorentz invariant, so the question still stands. To answer the question, we need to address two things: ... 2 A super-operator is a linear operator on \mathcal L(\mathcal H). Its eigenvectors are elements of \mathcal L(H). If you want to use matrix algebra, you have to pick a basis for \mathcal L(H), compute the matrix of the superoperator in this basis, then diagonalise it. The d^2\times 1 eigenvectors you find actually represent elements of \mathcal L(H) ... 2 To make this completely rigorous, you would have to delve into functional analysis. But I will try to explain it intuitively. (i) Arguably all derivatives of \psi are continuous in reality. Non-smooth functions are an idealization, but a useful one. They can be formalized using distributions. Even when f is discontinuous, f' is defined to be something ... 2 Maybe a more conventional way to see this: Write in polar form \alpha=\vert\alpha\vert e^{i\phi_a}, \beta=\vert\beta\vert e^{i \phi_b}. Then your state is$$ e^{i\phi_a}\left( \vert\alpha\vert \vert 0\rangle + e^{-i(\phi_b-\phi_a)}\vert \beta\vert\vert 1\rangle\right)\, . $$You can eliminate the overall phase e^{i\phi_a} as two states differing by an ... 2 As noted in Mike Stone's answer this is explained in any quantum mechanics textbook. The inner product is a complex inner product, in general:$$\langle \phi|\psi\rangle\in \Bbb C \tag{1}.$$In quantum mechanics the probability - given a system in the state |\psi\rangle - of finding it after measurement in the state |\phi\rangle (which will be an ... 2 The answer is "no". To this end, note that every U=M^TM has the property that$$ U^T = (M^TM)^T = M^TM = U\ , $$i.e. U is its own transpose. However, there are clearly U which are not of this form, such as$$ U = \frac{1}{\sqrt{2}}\begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix} (that is, a \pi rotation about the y axis). (Note ... 2 In the Dirac notation you use, one can express the initial state vector as $$\left|\Psi_0\right> = \sum_i a_i\left|\psi_i\right>$$ where the \left|\psi_i\right> are the eigenvector solutions. You then operate on the left of this with e^{-iHt/\hbar}, which is an operator (think of the series expansion of this into terms ... 2 The process is recursive. Start with \begin{align} a(\omega_1)a^\dagger(\nu_1) &=\delta(\omega_1-\nu_1)\mathbb{1}+a^\dagger(\nu_1)a(\omega_1)\, , \tag{1}\\ a(\omega_1)a^\dagger(\nu_1)a^\dagger(\nu_2) &=\delta(\omega_1-\nu_1)a^\dagger(\nu_2)+ a^\dagger(\nu_1)\left(a(\omega_1)a^\dagger(\nu_2) \right)\, , \tag{2} \end{align} and now use Eq.(1) to ... 1 You are introducing \psi_n(x) as the wavefunctions of the harmonic oscillator. But more precisely, they are the energy eigenfunctions having energy E_n = \hbar \omega \left ( n + \frac{1}{2} \right ). This means they evolve the way all energy eigenfunctions evolve which is $$\psi_n(x, t) = e^{-i E_n t / \hbar} \psi_n(x, 0).$$ ... 1 This is best understood with a specific example rather than a general mess of indices. Take H=\hbar\omega \sigma_z=\hbar\omega \left(\begin{array}{cc}1&0 \\ 0 &-1\end{array}\right)$and$\Psi(0)=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1\\ 1\end{array}\right)=\frac{1}{\sqrt{2}}\vert +\rangle +\frac{1}{\sqrt{2}}\vert -\rangle$, where$\sigma_z\...

1

Let us represent the state vectors as $\left| \psi_0 \right> = a\left| \phi_1 \right> + b\left| \phi_2 \right> \equiv \begin{pmatrix} a \\b \end{pmatrix}$ and $\left| \psi_1 \right> = c\left| \phi_1 \right> + d\left| \phi_2 \right> \equiv \begin{pmatrix} c \\ d \end{pmatrix}$, where $\left| \phi_1 \right>$ and $\left| \phi_2 \right>$ ...

1

RulerOfTheWorld, Both interpretations are possible, yet the second violates the principle of locality , since the effect of A on B is instantaneous. Therefore, your first interpretation is most likely the correct one since it is in agreement with the rest of physics, like relativity.

1

I had a physics teacher who used to say energy conservation will never be refuted because we'd just add an extra term that absorbs the apparent violation. That sounds like an accusation of unfalsifiability, but it makes sense because energy really does come in multiple forms that are detected in different ways, with varying degrees of difficulty. "Let's ...

1

Your notes are trying to say that when you have $$\psi(x) = \int_{-\infty}^{\infty} dx K(x,x^\prime) \psi(x^\prime)$$ for any reasonable function $\psi(x)$, the only possibility is that $K(x, x^\prime) = \delta(x - x^\prime)$. They "prove" this by considering the special case where $\psi(x)$ is a Dirac delta itself. (...

1

First, notice that if $L_y$ is continuous for every $y \in \mathcal H$, by the uniform boundedness principle you would conclude that $A$ is a bounded operator. So if $A$ is unbounded, there will be some $y\in\mathcal H$ such that $L_y$ is unbounded. Now, for any $y\in \mathcal H$, we know $L_y$ is at least defined on $\mathcal D(A)$. It can be bounded, in ...

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