# Tag Info

94

A "perfectly efficient" computer can mean many things, but, for the purposes of this answer, let's take it to mean a reversible computer (explained further as we go). The theoretical lower limit to energy needs in computing is the Landauer Limit, which states that the forgetting of one bit of information requires the input of work amounting $k\,T\,\log 2$ ...

9

There is no contradiction. This: "The de-Broglie wave and the particle are the same thing; there is nothing else. The real particle found in nature, has wave properties and that is a fact." is a more general statement. Note that it does not define what is "waving". It just states that the particle is characterized by a wave. This goes into the ...

7

Not every tensor is a simple tensor. The simple tensors of the form $a\otimes b\in A\otimes B$ span the tensor product, which means that a general element $t$ of the tensor product is a linear combination of these simple tensors, i.e. $$t = \sum c_{ij} a_i\otimes b_j$$ for some basis $a_i,b_j$ of $A$ and $B$ respectively. If $A$ itself is a space of ...

6

No. Counterexample: choose a basis in 2D quantum space, so the basis states are written $\left(\begin{array}{c}1\\0\end{array}\right)$ and $\left(\begin{array}{c}0\\1\end{array}\right)$. Now think of a mixture, with a state's having probability $p$ to be in state 1. The density matrix is then $\mathrm{diag}(p,\,1-p)$, which is not singular for ...

6

Group 2 has some non-objectable contents (“for these other particles, we do not even have a seemingly concrete macroscopic property to associate with the wave”), but is otherwise inconsistent (“We arrive at the conclusion...”: how does the “conclusion” relate to the previous statement in any way?) and wrong in the main aspect with which you're concerned ...

5

The answer to your question is NO. The simplest counter-example is the identity matrix. There is no solution to $|a\rangle\langle b| = I$. For example, try to solve $\begin{bmatrix} x \\ y \end{bmatrix} \begin{bmatrix} z & t \end{bmatrix} = \begin{bmatrix} 1&0 \\0&1 \end{bmatrix}$. Since $xt=0$ we know that $x=0$ or $t=0$. But clearly $x$ ...

5

Phenomena in quantum mechanics may be expressed using any basis. It doesn't mean that all bases are equally useful for a given situation. In particular, a fundamental postulate of quantum mechanics says that right after every measurement, the system is found in one of the eigenstates of the observable that was just measured. That's why the basis of the ...

5

Is time made into one observable? No. It is known that an operator $T$ that satisfies $[H,T]=i\hbar$ is either self-adjoing and $H$ unbounded below or anti-self-adjoint. Therefore, the theory is either intrinsically flawed (arbitrary negative energy) or $T$ is not observable (anti-self-adjoint $\Rightarrow$ imaginary eigenvalues).

5

For the absolutely continuous part of the spectrum of a self-adjoint operator $H$, the "density of states" is provided by the Radon-Nikódym derivative of the spectral measure of $HP_{ac}$ with respect to Lebesgue measure, where $P_{ac}$ is the orthogonal projection onto the absolutely continuous subspace of the domain of $H$. This formula is well defined ...

4

The basis is still $\{|\boldsymbol r\rangle\}$. The abstract Schrödinger equation is $$i\frac{\mathrm d}{\mathrm dt}|\psi\rangle=H|\psi\rangle$$ where $|\psi\rangle$ is a set of four kets, (with a slight abuse of notation) $$|\psi\rangle=\begin{pmatrix}|\psi_1\rangle\\|\psi_2\rangle\\|\psi_3\rangle\\|\psi_4\rangle\end{pmatrix}$$ Time is still a ...

4

I am not sure I understand perfectly you question but formally in the canonical ensemble we can write the partition function $Q(\beta)$ as being: $$Q(\beta) = \int\cdot \cdot \int d\mu(x)\: e^{-\beta H(x)} = \int_0^{+\infty} dE \: \rho(E)e^{-\beta E}$$ where $d\mu(x)$ is the volume measure for the micro states in the system, ...

4

Due to the exponential dependendence of tunneling probability on barrier height and thickness, it is entirely possible for tunneling to take femtoseconds, or for tunneling to take 100 trillion years. The point is that tunneling can potentially be very fast (femtoseconds), and those devices where tunneling is supposed to happen are designed so that tunneling ...

4

Historically, you probably want to start with the de Broglie relations (i.e. $p = \hbar k$), which are just a wild guess. This immediately pops out the form of $p$ as an operator if the wavefunction is a plane wave. Mathematically, $p$ should be defined as the generator of translations (or equivalently the conserved quantity corresponding to translational ...

4

In the quantum mechanical description of any physical system, including a quantum field or a collection of interacting quantum fields, there is always one state vector – one collection of numbers (probability amplitudes) that generalizes what is referred to as the "wave function" in quantum mechanics of particles. In quantum field theory, a better name is a ...

3

For your initial check, you used the formula $c = \nu \lambda$, giving the incorrect formula $E = pc$. This is wrong, matter waves don't travel at the speed of light; instead, $\nu \lambda$ is equal to the phase velocity $v_p$. You then use $E = pc$ in all the other derivations, making them all wrong. Hypothesis #1 is correct for relativistic quantum ...

3

OP explicitly asks whether a material object (i.e. not a state of light) has been placed in the superposition $|\psi⟩=|\psi_1⟩+|\psi_2⟩$, where $|\psi_n⟩$ is the $n$th Fock state of a harmonic oscillator. Perhaps the clearest example of this is the achievement of precisely that superposition in the quantum state of a cantilever microwave resonator ...

3

First, there's no physical $c\to 0$ limit of relativity because that limit basically says $v\gg c$, velocities are much larger than $c$. But relativity prohibits $v\gt c$, let alone $v\gg c$, so this limit $c\to 0$ i.e. $v\gg c$ cannot be taken. Relativity generalizes non-relativistic physics so that it allows large $v$, $v\sim c$, but at the same moment, it ...

2

If you are happy with Bob being right only in some percentage of the cases, there is a much easier protocol which does not even require entanglement: Just let Bob guess Alice's bits. He will be right in 50% of the cases. Note that once the probability to guess the right result is above 50%, communication is possible, e.g. by doing majority voting. In ...

2

These relations are found in every book on QM, but the usual notation is $$X|x\rangle=x|x\rangle$$ and $$P|p\rangle=p|p\rangle$$ To go from these equations to the ones you've written, you just have to project them into the position basis $|x'\rangle$ (and use $\langle x'|x\rangle=\delta(x-x')$ and $\langle x'|p\rangle\sim\exp[ipx]$). Edit Important: ...

2

I'm just guessing here, but one way to interpret it goes like this. If you have the amplitude $\Phi$ for an outcome that can be achieved in multiple indistinguishable ways each described by an amplitude $\phi_i$ with $i \in 1, 2, \dots , n$ we write $$\Phi = \sum_{i=1}^n \phi_i \,.$$ That makes the probability for the outcome \begin{align*} P &= ...

2

Short answer: no. I'll give some context with the details of the simplest examples. In the context of conservation laws, "energy" refers to the Hamiltonian. In classical mechanics, a quantity without explicit time dependence is conserved iff its Poisson bracket with the Hamiltonian is 0. In quantum mechanics, quantities are promoted to operators on a ...

2

I agree with CuriousOne that you would be better off ditching both of these viewpoints and looking for something more modern. However, this is instructive because it does illustrate a common problem in QM education: many authors are invested in a particular interpretation, and present that interpretation (disingenuously, to my thinking) as the only correct ...

2

It's neither a classical wave nor a classical particle. I think any attempts to describe it as either of those need to be qualified like this. It might look like one or the other, but both are only approximations. The best theories we have describe quantum fields, and a particle is a field quantum. I don't really know how to describe a field quantum in ...

2

Every complex number can be written in the form $re^{i\theta}$ for a real number $r$. We call $e^{i\theta}$ the phase. For example, if $$|\psi \rangle = \frac{1}{\sqrt{2}} ( |0 \rangle + i |1 \rangle)$$ then the phases of the $|0 \rangle$ and $|1 \rangle$ components are $1$ and $i$, and their relative phase is $i$. Now consider $$|\psi' \rangle = ... 2 The argument is that is a particle has lover mass it will be more easily effected by external forces. Since the electrons are ~1000 times lighter than atomic nuclei it means that they also move on time-scale that are ~1000 faster. Thus any change to the nuclear positions will be very slow from the point of view of the electrons and so one can assume that ... 2 It seems I have figured out an answer for 2 terms in the original state. Suppose that the state is$$\rho = a |\alpha \rangle \langle \alpha | + (1-a) |\beta\rangle \langle \beta|$$We need to write it in a basis, which is$$|+\rangle = \frac{|\alpha\rangle + |\beta\rangle}{\sqrt{2}}; \quad |-\rangle = \frac{|\alpha\rangle - |\beta\rangle}{\sqrt{2}}. $$... 2 The short answer is that only if the Berry curvature is defined by: (in matrix notation):$$F_{\mu \nu} = \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu} + [A_{\mu}, A_{\nu}]$$it becomes gauge covariant, i.e., for a gauge transformation:$$A_{\mu} \rightarrow g^{-1}A_{\mu} g+g^{-1}\partial_{\mu}g $g \in U(N)$ ($N$ is the degeneracy of the level), the ...

2

Spontaneous emission and stimulated emission are cases related to measurement, but they are not identical to measurement. We might ask how the measurement of states and the preparation of states are related to measurement, or the decoherence of states. Both are necessary aspects of quantum physics. In order to do experiments we need to prepare identical ...

2

Engtangled photons: first you must understand that the photon is the particle obtained when the modes of the electromagnetic field are quantized, and that they are created and destroyed as discrete quanta of energy, in agreement with Planck's relation, $E=hf$, where $f$ is the frequency of the electromagnetic field corresponding to the quantized mode; that ...

2

I thought I would indicate how radiactive decay can be argued to occur from quantum tunneling. This derives in a very “back of the envelope” way the phenomenological equation for radioactive decay. From there I can argue some about the role of observing radioactive decay. Quantum tunneling can be see with the square potential barrier. A square potential ...

Only top voted, non community-wiki answers of a minimum length are eligible