# Tag Info

## New answers tagged quantum-mechanics

0

The wavefunction $\psi(x_0, x_1, t)$ gives the probability of finding one electron between $x_0$ and $x_0+dx_0$ and the other between $x_1+dx_1$: $$P(x_0, x_0+dx_0; x_1, x_1+dx_1; t) = \vert\psi(x_0, x_1, t)\vert^2dx_0dx_1$$ We expect the brightness of the scintillation at $y$ to be given by the probability (density) that either electron is found at $y$: ...

0

For any arbitrary collection of such travelling waves will always be a wave envelope that retains the same shape as the collection of waves propagate? No, it will not. For example, a Gaussian wave-packet will spread out in time. Wave packets are used to represent localization of particles in Quantum Mechanics.Group velocity will give the physical velocity of ...

-1

I did not learn the quantum theory on a relativistic view. It's true that it's a hard topic to understand thats why it may look so relative, but as I was told it isn't, we just don't have all the informations we need about it. You can work on yes/no experiments. In some books writes that light behaves like a wave AND like a particle, but in fact wave is a ...

2

Let's take the canonical commutation relations (CCR), in their exponentiated form (Weyl's relations): $$V(\eta)T(q)=e^{-i\eta\cdot q}T(q)V(\eta)\; ,$$ where $\{V(\eta)\}_{\eta\in \mathbb{R}^d}$ and $\{T(q)\}_{q\in \mathbb{R}^d}$ are objects of a given normed algebra with involution. This is a very general notion, that is nowadays taken as the definition of ...

0

Let me attempt an answer, based on the comments from Timaeus and ACuriousMind.. Answer: No, there is nothing that forces us to view non-relativistic quantum mechanics as a $U(1)$ gauge theory. It's an elegant, equivalent way of looking at some situations like the Aharanov-Bohm effect. In more detail: We could treat the Aharanov-Bohm effect in the usual, ...

1

Time varying intensity means time varying flux of photons. Above threshold this gives a consequent time varying photocurrent at anode. This is the basis of optical sound since the 1920's, where the intensity of light was modulated by the varyng transparency of the soundtrack on the film and then read by a phototube. Below threshold: as already answered, very ...

0

here's a short simple answer: the Hamiltonian of the pencil can be approximated by an inverted harmonic oscillator near the equilibrium (downward parabola) . It's an easy exercise to solve.

0

In an one dimensional harmonic oscillator the energy observable is a complete set of observables of the system (we don't have degeneracy). If you have a list of possible energies you know everything about that system when we don't have degeneracy. If this list is an infinite, we have precise the list of energies of an harmonic oscillator. You can construct ...

0

Things are amazing for as long as you do not (or refuse to) understand them! What happens is that, counterintuitively, everything around us is waves (matter or de Broglie waves). Hence they naturally interfere, producing for example the diffraction patterns behind a double slit (or indeed even a single slit). Waves naturally and unsurprisingly obey the ...

1

If the particle source is "unpolarized", that literally means it is equally likely to find particles from this source in either energy eigenstate - that's the definition of "unpolarized", so you shouldn't be surprised about that. When the spin of a particle is "perpendicular to the magnetic field", that's another way of saying that the particle is in an ...

1

The energy difference along with the larger thermodynamical likelihood for occupation of the lower level is real. There is an application, nuclear magnetic resonance (NMR) spectroscopy/imagining/quantum computing. But due to the very small energy difference for technically achievable magnetic fields, the effect is usually negligible at roomtemperature. NMR ...

6

Any "harmonic oscillator", seen as the second quantization operator $$d\Gamma(1)=\int a^*(k)a(k)dk$$ of the symmetric Fock space $\Gamma_s(\mathscr{H})$ over a (separable) Hilbert space $\mathscr{H}$, has the natural numbers $\mathbb{N}$ as spectrum (i.e. evenly spaced spectrum). In addition, if e.g. $\mathscr{H}=\mathbb{C}$, the operator $aa^*$ has ...

2

If we take two slits large enough for bowling balls and with a much smaller than the ball separation , and a pile of bowling balls and design a catapult throwing the balls at the slits parallel to the ground within the window covering the two slits, what will happen? 1) some balls will pass through one of the slits without touching, straight ahead 2)some ...

0

Are you curious about the case where the $a_j, b_k$ are matrices or operators? As in gautam1168's answer. $$(\mathbf{a}\times \mathbf{b})_i = \epsilon_{ijk} a_j b_k$$ So taking the dagger: $$(\mathbf{a}\times \mathbf{b})_i^{\dagger} = (\epsilon_{ijk})^\dagger b_k^\dagger a_j^\dagger,$$ because $(\epsilon_{ijk})^\dagger$ is just a number which commutes with ...

0

Simply to say, just abandon the terms corresponding to the part on which the partial trace is applied. For example, $$\rho = \sum_i c_i |A_i,B_i\rangle \langle A_i,B_i|,$$ then $${\rm Tr}_B[\rho] = \sum_i c_i |A_i\rangle \langle A_i|,$$ while $${\rm Tr}_A[\rho] = \sum_i c_i |B_i\rangle \langle B_i|.$$ In your case, $i=1,2$, $A_1 = A_2 = g_A$, $B_1 = ... 0 The partial trace with respect to$B$is a map from the operators on the hilbert space$A \otimes B$to the operators in$A$given as follows$\langle \phi | Tr_B (\rho_{AB}) |\psi \rangle = \sum_n \langle \phi| \langle n | \rho_{AB} | \psi \rangle | n \rangle$Where$\psi,\phi$are vectors in A, and$|n\rangle$is a basis in B. So, in your example: ... 1 Further to Anna V's answer, in the case of an electron there is an important physical meaning to the "lack" measurability of the phase of the electron's wavefunction. This is because the electron is coupled to the electromagnetic field. And, if one models this by the Minimal Coupling between the electron and the electromagnetic field, one gets the ... 3 What does measurable mean ? It means that one can do an experiment and get a value for a+ib , the complex number. A complex number to be measurable one should be able to measure a value at the same time for a and b and put a point on the complex plane. This means two independent variables, a and b can be measured and a point defined. In quantum mechanics ... 0 I want to add, that the situation changes if the elements of these vectors are operators. Since this is what your notation might imply. Taking the definition of what a vector product for these quantities should mean, i.e. $$(\mathbf{a}\times\mathbf{b})_k = \epsilon_{ijk} a_i b_j \hspace{1cm} (1)$$ then gives by definition (note that$(AB)^\dagger = ...

1

The slater determinant is only a "trick" to get a total antisymmetric wave function. This is required by the Pauli principle. For understand this you need to think in indistinguishability of particles. So a any allowed state of a particle need to be assigned equally at each indistinguishable particle in your system. So if you have to state allowed: ...

0

As ACuriousMind said your expression is not correct: $$(\mathbf{a}\times \mathbf{b})_i = \epsilon_{ijk} a_j b_k$$ So taking the dagger: $$(\mathbf{a}\times \mathbf{b})_i^{\dagger} = (\epsilon_{ijk})^\dagger a_j^\dagger b_k^\dagger\\ =\epsilon_{ijk} a_j^* b_k^*$$ because $\epsilon_{ijk}$ is real scalar, and $a_j$ and $b_k$ are complex scalars. (for scalars ...

2

First I want to point out that most of these questions do not bring up issues specific to Bohmian mechanics. That's not a criticism, I'm just pointing out that these notations and concepts are already employed in standard quantum mechanics, or even in classical mechanics. I am going to answer this a little casually, and then make my answer "community ...

0

Ψ is supposedly a probability density amplitude. ΨΨ* is the probability density which in theory can be measured. For example in electron diffraction through a crystal a statistical measure of the electrons in, divided into the electrons out in a small region divided by the volume would allow ΨΨ* to be approximately measured. Phase information is lost when ...

1

It actually is the very essence of the QM. In short, when we observe a superposed state, the probability of observing specific eigenvalue is the square of the norm of the corresponding eigenstate in the superposed state. And this is more like a postulate, rather than a mathematical derivation. For example, particle in a box has discrete eigenvalues, bounded ...

1

Eigenstates aren't the only allowed physical states. It's a postulate of quantum mechanics that the most general quantum state can be written as a superposition of eigenstates of some operator (the Hamiltonian for instance). For instance $\Psi(x)=\sum_nc_n\psi_n(x)$ is a general quantum state for a particle in a box, where $\psi_n(x)$ are the energy ...

1

Does $$L_+ L_- Y_{lm}$$ ,where $Y_{lm}$ is a spherical harmonic function, equals to zero. If so, why? It may or may not equal zero depending on the value of $m$. If $m$ is equal to $-l$ then yes, otherwise no. If $m=-l$ then applying the lowering operator annihilates the state (i.e., gives zero) since there is no state with an $m$ lower than $-l$. ...

2

A magnetic field consists of photons. Photons are spin $1$ particles, which means for a given $z$ axis a measurement of spin can yield $+1$ or $-1$. If a photon collides with an electron, we know that spin must be conserved. Let's assume the electron is in a spin up state $\uparrow_e= + \frac{1}{2}$ and the photon in a spin down state $\downarrow_P= -1$. ...

3

$\vert+\rangle$ and $\vert-\rangle$ are really just shorthand notations for the two eigenvectors of the diagonal spin operator $\sigma_z$. This means concretely: $$\vert+\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$ $$\vert-\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$ Therefore the action of the sigma operator gives you simply the corresponding ...

2


3

The classical limit of quantum theories is understood quite well from a mathematical standpoint nowadays. The so-called semiclassical analysis covers the QM (finite dimensional phase-space) cases, the Hepp method and infinite-dimensional semiclassical analysis cover the systems with classically infinitely many degrees of freedom. The ideas can be summed up ...

2

If interpretations actually had different physical consequences, we might go about calling them different 'theories' instead. For example, one might say that the classical Newtonian model of gravity comes from many non-interfering (straight line) threads of variable length, maybe with complex infinitesimal modern art on them, (in fact, undetectable in any ...

0

Different interpretations may have different physical meanings. For the most part they probably act to "translate" the reality into something intuitive for the person, but some may make unique predictions that just aren't tested yet and we cannot yet know if the interpretation is accurate.

-1

If I have a model of the universe with all the laws in a computer program. Can I say that the electrons that are modelled in the program exist? No. The map is not the territory. In such a way I am not shocked if someone describes to me an experiment that shows "as if" an electron goes through two slits at the same time. It goes through two slits at the ...

5

The answer is that the premise is wrong. There can't be a hydrogen wave function with the coefficients you have written. Even if there was no $| 1 0 0 \rangle$ state present, the state isn't normalized. That means that it isn't physical. However, remember that the coefficients are somewhat arbitrary, that is, we're allowed to multiply the whole wavefunction ...

1

Looks like textbook hybridization problem, did you check the usual suspects, or e.g. this one?

2

The state you have given is not normalisable as a consequence of the results of the calculations you have done. Even if the first state (with coefficient $A$) was not present it would not be normalised. To normalise what you have given, another constant needs to multiply everything through (so that the relative proportions are unchanged

0

The spin of a particle flips under time reversal. It has nothing to do with the phase of the particle! You may go through time reversal symmetry for more information about spin flipping.

1

Comments to the question (v2): Normally we assume that $-\infty< x_1 < x_2<\infty$ are the turning points in the 1D potential well. This means $$\forall x\in ]x_1,x_2[:~~ E ~>~ V(x).$$ Hence OP's third last equation $$0 ~<~ \int_{x_1}^{x_2}{dx\ \sqrt{2m(E-V(x))}} ~=~ -\frac{\pi}{2}\hbar~<~0$$ can never be fulfilled. The correct WKB ...

0

Apart from (physically motivated) asymptotic prescribed behaviour at infinity $|x|\to\infty$, we don't actually impose/demand/require any conditions on the wave function $\psi$ beyond the TDSE. Continuity and (possibly higher) smooth conditions are instead derived from a standard bootstrap argument, see e.g. my Phys.SE answer here for details.

Top 50 recent answers are included