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8

This is an important question. You are correct that the energy expectation values do not depend on this phase. However, consider the spatial probability density $|\psi|^{2}$. If we have an arbitrary superposition of states $\psi = c_{1} \phi_{1} + c_{2} \phi_{2}$, then this becomes $|\psi|^{2} = |c_{1}|^{2}|\phi_{1}^{2} + |c_{2}|^{2} |\phi_{2}|^{2} + ...


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 ...


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 ...


4

Observe that $$ p \langle x \vert p \rangle = \langle x \vert \hat{p} \vert p \rangle =-\mathrm{i} \partial_x \langle x \vert p \rangle$$ since, with $\hbar = 1$, $\hat p = -\mathrm{i}\partial_x$ when acting on wavefunctions $\langle x \vert \psi \rangle$. The solution of this differential equation is then the inner product $$ \langle x \vert p \rangle = ...


4

The question doesn't have an answer because electrons aren't waves and they aren't particles. This is a common source of confusion, and has led to the endless debates about wave particle duality. Quantum systems are described by a wavefunction that can behave as a wave in some circumstances and behave like a particle in others.However it is vital to ...


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 ...


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 ...


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 ...


3

There are - even after clarifications in the comments - several ways to interpret the question. So let me start out with the way that fits all parameters here: Suppose we have a Hilbert space $\mathcal{H}$ (for qubits, $\mathcal{H}=\mathbb{C}^2$) and suppose we take a (pure) state in the Hilbert space $\mathcal{H}\otimes \mathcal{H}$. Question 1: If we ...


3

Also you can modify the wavefunction with a global phase $\psi(x)\rightarrow e^{i\phi}\psi(x)$ without affecting any expectation values because the phase factor will cancel when taking inner products, so this global phase doesn't contain any information. Only relative phases are meaningful in quantum mechanics.


3

Mathematical reason is that the time evolution operator is unitary, which means that $U^\dagger = U^{-1}$. Therefore $\langle \psi(t) | \psi(t) \rangle = \langle \psi(0)| U^\dagger U | \psi(0) \rangle = \langle \psi(0) | \psi(0) \rangle$. We can see that it is unitary by considering the Schrodinger equation: $$ \newcommand{\ket}[1]{| #1 \rangle} ...


2

Application 1: It's the $z$-component of the vector valued angular momentum observable for a spin $\frac{1}{2}$ particle, when the basis states are the $z$-component angular momentum eigenstates. If this sounds a bit circular and tautological, it is the reason why $\sigma_z$ is diagonal. So the $n^{th}$ moment of the probability distribution of an angular ...


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 ...


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


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 ...


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 $\downarrow_P= -1$. ...


2

How to prove it? EDIT: Consider a state $|\psi_n\rangle$, which is an eigenstate of the Hamiltonian $H$ having eigenvalue $E_n$. Then consider the action of $H$ on these two other states: $$ |\alpha\rangle \equiv AB|\psi_n\rangle $$ and $$ |\beta\rangle \equiv BA|\psi_n\rangle $$ EDIT: Ask yourself: Are these eigenstate of the Hamiltonian? If so, ...


2

You might see it from the notion of Fourier transformation. For example, you could express an arbitrary quantum state of your Hilbert space in momentum representation by applying a Fourier transformation on your position representation. More explicitly, for $\left|\psi\right>\in\mathcal{H}$, you might define $$\left< \mathbf{q} |\psi\right> = ...


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 ...


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 ...


1

I am not supposed to place an answer here, as I am no more active in this site: however, a comment doesn't offer enough place. So, you ask: 1) "spin is a property of the wave function, and not of the particle?" Please pay attention to the following differences between the standard quantum theory (SQT) and the Bohmian interpretation (BI): SQT ...


1

Here's a way to get a decent distance estimate for solids, liquids or gases.: for solids or liquids, you can get the number density, $n$, of atoms or molecules (as needed), from the density, Avogadro's number ($6.02E23$) and the molecular weight ($\rho,N_a,W$: $$n = \frac{\rho N_a}{W}$$ for a gas with pressure, temperature and the boltzmann constant ...


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 $-l$. ...


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

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

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: ...


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 ...


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 ...


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

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 ...



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