# Tag Info

1

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: ...

0

Let assume two states $|x\rangle$ and $|y\rangle$ and their inner product $\langle x|y\rangle$. If $\langle x|y\rangle = 0$, then both state can be perfecly distinguished and there is no uncertainty which is which. Examples of such states are $|0\rangle$ and $|1\rangle$ or $|+\rangle$ or $|-\rangle$. As $\langle x|y\rangle$ tends to $1$, the states are more ...

0

In the abstract of this paper they say The algorithmic output is distinguishable from noise, in contrast to previous demonstrations. The previous demonstrations seem to be claims to have factored $15$. Shor's algorithm works (slowly) even if your "quantum" computer has a coherence time of zero. The reason is that it runs the quantum subroutine in ...

1

The electron shells surrounding an atom possess progressively higher energy levels as their corresponding quantum numbers are indexed upwards. Roughly speaking, as you add build larger and larger nuclei, the bigger the nucleus (that is, the more protons you populate it with), the larger the the resulting electron cloud becomes and the more oddly-shaped and &...

1

Yes, such superpositions are allowed. Relativistic quantum field theory wouldn't work without them. Consider an initial state $|n\rangle$ with just a single isolated neutron. That's a single-fermion state. Over the next several minutes, in the Schrödinger picture, the single-neutron component of the state smoothly declines from $1$ toward $0$, and the proton-...

1

Eigenstates of the 1-D infinitely deep square well form a complete basis in the Hilbert space. Thus, given the particle's wave function, there's only one way to write $\Psi(x)$ as a linear combination of the relevant eigenstates. Especially, if you consider the Fourier expansion of the function $\Psi(x)=Nx$, you'll realize that the particle must have been ...

0

The lectures are dismissing boundary terms (BTs) in the action. This is strictly speaking not correct. In fact, BTs are important for the possible compatible choice of boundary conditions, cf. e.g. this related Phys.SE post.

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 ...

1

Regarding (i), I am still not sure what you are asking. In general, $d\psi/dx$ is continuous over an interval if the potential $V(x)$ is finite within that interval. At the boundaries, if the potential $V(x) \to \pm \infty$, then $d\psi/dx$ is no longer continuous. Therefore, the continuity of the first derivative of the wavefunction depends on the behavior ...

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 ...

0

The energy of a crystalline solid structure depends on the type of interatomic interaction and a delicate trade-off between distances and symmetry of the crystalline structure. Moreover, one has to consider that at the freezing point, finite temperature effects (via entropic contribution) may stabilize some crystalline structures that are not the most stable ...

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 ...

0

Why not evaluate this in coordinate representation? \begin{align} \langle n,l',m'|z|n,l,m\rangle&=\langle n,l',m'|\left(\int d^3r''|\mathbf{r''}\rangle\langle \mathbf{r''}|\right)z\left(\int d^3r'''|\mathbf{r'''}\rangle\langle \mathbf{r'''}|\right)|n,l,m\rangle \\ &=\left(\int d^3r'' \langle n,l',m'|\mathbf{r''}\rangle\right) \left(\int d^3r'''\... 0 I am liking the answers which have been posted, but figure to offer an answer which uses more primitive means. This system has a classical analog with Lagrangian of the form: L=\frac{1}{2}\cdot \dot{x}\cdot M\cdot \dot{x}-\frac{1}{2}\cdot x\cdot K\cdot x\mp \frac{1}{2}\cdot \dot{x}\cdot C\cdot x $$I say \mp \frac{1}{2}\cdot \dot{x}\cdot C\cdot x since ... -1 The second method is correct. This is because (a_1+ib_1\ 0)^T is not the matrix of the wavefunction \psi_1. Rather-$$\psi=(a_1+ib_1)\psi_1+(a_2+ib_2)\psi_2$$In matrix form when you wrote the expression for \psi, you already assumed that \{\psi_1,\psi_2\} is the basis. Thus the matrix you have written for \psi above is the coordinatisation of \... 0 In my view this characterization naturally appears when one studies LSZ Reduction. Let us label the unitary irreps of the universal cover of the Poincaré group by (m,s) where m\in [0,+\infty) is the mass and s\in \frac{1}{2}\mathbb{Z} is the spin/helicity. Suppose you have some particle in the irrep (m,s) whose state space is spanned by the improper ... 1 An inner product (more specifically a positive definite inner product) defines the notions of length of a vector and projection/angle of one vector along another. While length can easily be real and positive$$\langle \psi | \psi \rangle \in \bf R^+$$which follows from the properties of the inner product, the inner product between two different vectors ... 2 As noted in Mike Stone's answer this is explained in any quantum mechanics textbook. The inner product is a complex inner product, in general:$$\langle \phi|\psi\rangle\in \Bbb C \tag{1}.$$In quantum mechanics the probability - given a system in the state |\psi\rangle - of finding it after measurement in the state |\phi\rangle (which will be an ... 1 I'm not sure why you'd expect \langle \phi|\psi\rangle to be equal to 1. In elementary mechanics, we deal with the vector spaces \mathbb R^2 and \mathbb R^3. If you have two vectors \vec a and \vec b, \vec a \cdot \vec a = \vec b \cdot \vec b = 1 means that \vec a and \vec b both have magnitude 1, but if I gave you no other information ... 1 Was the double slit experiment tried with slow light? Slow light means that the group velocity of light is slowed down to the value you give. The experiments with slowing the velocity are done in an apropirate medium , not in space where the double slit single photon experiment, for example, is done. This shows a wave with the group velocity and phase ... -1 The physical quantity |\langle \psi|\phi\rangle|^2= \langle \phi|\psi\rangle \langle \psi|\phi\rangle is real. Its interpretation is explained in any Quantum Mechanics textbook. 0 Note that, for any pair of matrices A,B, you have$$e^A B e^{-A} = e^{{\rm ad}(A)} B \equiv \sum_{k=0}^\infty \frac{1}{k!}[\underbrace{A,[A,\cdots ,[A}_k,B]\cdots]] \equiv B + [A,B] + \frac12 [A,[A,B]] + \dots,$$where {\rm ad}(A) denotes the adjoint operator, {\rm ad}(A):B\mapsto [A,B], and the complicated-looking object in the series is a repeated ... 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 ... 1 I have not heard of the exact experimental setup you illustrate, but the experiment has been done by passing electrons, one-at-a-time, through a double slit. The single-electron double split experiment was carried out by Stefan Frabboni and coworkers and is mentioned in this IOP article. The predicted interference pattern was observed. Edit: —————————————————... 0 When a single photon and a coherent state are incident on the input modes "b" and "a" of a beam splitter, there are four possible outcomes at the output. Both the inputs are fully transmitted(this is called photon catalysis). The condition for this to happen is to have a beam splitter with a very high transmittance. The reflectance and ... 0 Let's say we only need the first order corrections. Do we need to diagonalize the degenerate subspace of the perturbation? Yes, in order to apply the perturbation theory we need the perturbation matrix to be in a diagonal form, in this way we can properly calculate the energies of the multiplet. This statement does not mean that we must always diagonalize a ... 0 As you mentioned, photons in the box have quantised momenta based on the fact that they are standing waves, with an integer number of half wavelengths. \lambda = \frac{n}{2L} p_n = \frac{h}{\lambda} = \frac{2hL}{n} There are photons of all allowed momenta in the box. Which momenta are allowed is determined by the expression for p_n above, where n is ... 0 Black body radiation (BBR) results from the equilibriated thermal radiation that is emitted by the charges in a material as the material is heated. It is therefore a function of only temperature and is independent of material or geometry of the body, as also mentioned in the video you linked. Hence your question is answered in the negative. Yes the total ... 0 The pinholes would have the same colour if the cavities are at the same temperature. Imagine two different cavities, joined at their respective pinholes but isolated from the rest of the universe. These two cavities would eventually arrive at thermal equilibrium at the same temperature with no net flow of energy through the join. That is as expected because ... -1 I'm not sure if this will help but a very sharp former colleague once posed this question. You have two boxes. The first contains 1000 horizontally polarized photons and 1000 vertically polarized photons while the second contains 1000 left circularly polarized photons and 1000 right circularly polarized photons. How do you tell which is which? Since this is ... 1 You have stated case #1 incorrectly. Alice's 45 degree polarizer (in this case Anti-Diagonal) is not the same as Bob's 45 degree polarizer (in this case Diagonal). The two polarizers are perpendicular or opposite to each other. Therefore when a photon pair are ANTI-CORRELATED the coincidences should be at maximum because when one photon lines up with an ... 0 Let D:=d/dx. I'll give another proof D^\dagger=-D. using indefinite integrals (over an unstated space, such as \Bbb R or \Bbb R^3):$$\langle u|D^\dagger|v\rangle=\overline{\langle v|D|u\rangle}=\overline{\int v^\ast Dudx}=\int v Du^\ast dx=-\int u^\ast Dvdx=-\langle u|D^\dagger|v\rangle,$$where the penultimate = uses integration by parts (its ... 1 Note that :$$\left(\frac{d}{dx}\right)^\dagger=-\frac{d}{dx}a=(\cdots )X+(\cdots )\frac{d}{dx}a^\dagger=(\cdots )X-(\cdots )\frac{d}{dx}$$As expected a\not=a^\dagger. Edit: It's not a rigorous proof$$P=P^\dagger P\rightarrow -i\hbar \frac{d}{dx}-i\hbar \frac{d}{dx}=\left(-i\hbar \frac{d}{dx}\right)^\dagger=+i\hbar\left(\frac{d}{dx}\...

0

I have explored this topic myself in a couple of ArXiv papers The Concept of Entropic Time and Quantum Information and the Mind-Body Problem, so it interesting to know about the Page/Wooters theory and Quantum Magazine article, that I had not come across before. Thank you for that! To my mind, the proper framework for this discussion is decoherence theory, ...

1

The crucial insight made by quantum mechanics is that electromagnetic waves in the cavity can be described by simple harmonic oscillators. If the angular frequency of the mode of oscillation is $\omega$ (FIXED) , then the energy associated with this mode is given by $$E_n=\hbar \omega \left(n+\frac{1}{2}\right)\ \ \ \ \ n=0,1,2,\cdots$$ which are ...

1

The only relevant quantum number in this problem is the spin. Every particle is characterised by a state $|s, s_z\rangle$ in the Hilbert space $\mathcal{H}$. The first three questions can be answered by simple tensor product decomposition while the last one requires a simple trick to put the hamitonian in a better suited form.

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 ...

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, ...

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. 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. ... 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. 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 ... 0 whether virtual particles really exist or not, but if they did, would they be an exception to the law of conservation of energy? This is a simple Feynman diagram where a virtual particle is needed, in order to calculate the crossection to first order of electron-electron scattering. What exists are the two electrons interacting on the bottom, they exist ... 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. (...

Top 50 recent answers are included