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5

But how can we guarantee that two solutions $\boldsymbol {\psi_1}$ and $\boldsymbol {\psi_2}$ to the time-dependent equation don't have $\boldsymbol {\psi_1(x,0)} = \boldsymbol {\psi_2(x,0)}$. If we can't guarantee this, then how do we know that the solution found by Griffith's method is unique? I interpret that your question basically asks how do we ...


4

You're not missing anything. You are right, $k=\omega/c$. The argument $\sqrt{\frac{\omega ^2}{c^2}-k_z^2}$ in the Bessel function is the projection of the wavevector onto the radial direction. The use of Bessel functions beclouds what's going on a bit. Recall that a plane wave with wavevector $\vec{k}$ has the functional variation $\psi(\vec{r}) = ...


4

The definition of the expectation value of an operator A is \begin{equation} \langle A\rangle=\int{\psi^* (x) A(x) \psi (x) dx} \end{equation} (because it represents "the value of the variable" $A(x)$ times "the probability of being in that configuration" $P(x)=\psi^* (x) \psi (x)$) and for the particular case of the expectation value of the position ...


3

To get a wave function one has to solve the quantum mechanical equation for the boundary conditions of the experiment:"electron impinging on two slits". In the usual description one is using approximations : the incoming electron is a plane wave, the effective potential of the electrons in the matter of the slits is high. Then the distance between slits and ...


3

In general you need both kinds of functions $H_n^{(1)}(k\,r)\,e^{i\,n\,\phi} = (J_n(k\,r) + i\,Y_n(k\,r))\,e^{i\,n\,\phi}$ and $H_n^{(2)}(k\,r)\,e^{i\,n\,\phi} = (J_n(k\,r) - i\,Y_n(k\,r))\,e^{i\,n\,\phi}$ in any superposition, where $(r\,\phi)$ are the polar co-ordinates and $k$ the wavenumber. If the homogeneous region in question contains the point $r=0$ ...


3

What is a wave function? It is the solution of a quantum mechanical equation ( with the appropriate potentials),on which boundary conditions are imposed to make it specific to a system . $|\psi\rangle$ by itself is not independent of the environment the way that the operators X are. Thus the answer depends on the system under consideration. I like ...


3

The form of the solution shown by Griffiths is not unique. That means that there exist cases where a basis $\{\psi_n(x)\}$ will reproduce $\Psi$ as $$ \Psi(x,t)=\sum_{n=1}^\infty c_n\psi_n(x) e^{-iE_nt/\hbar}, $$ but there exists a second, different basis $\{\varphi_n(x)\}$ which (with different coefficients) also reconstructs $\Psi$: $$ ...


2

It must not be greater than 1. To find the probability function you must integrate the probability density, $\psi^* \psi$, over the region in which you want to calculate the probability: $$P_n(x_1,x_2)=\int_{x_1}^{x_2} \psi^*(x)\psi(x)\,dx \, .$$


2

No, if you observe which slit they traveled through then there is NOT an interference pattern. The act of observing, or more accurately, the need for the location of the electron to be resolved causes it to take on a definite position and then continue on from that position as a particle. If it is not observed or interacted with in some way that would make ...


2

The first 2-D image you posted is a typical simplification for teaching purposes. In it, they use the height of the sine wave to represent magnitude, and the directions of the sine waves to show how the fields point relative to each other. The light itself however is not itself at all cone-like. You have to imagine this sine wave existing at multiple points ...


2

A wavefunction of a quantum system is the system's state written in a particular form - no more nor any less than that. Here the word state has an exactly analogous meaning to the state of a classical system insofar that the state at any time uniquely defines the state at any other time and contrariwise. The state's evolution with time in either a quantum or ...


1

I now come to my point, why one restricts a particle's motion to some discrete set of distances? Is it to provide a theory on the particle's stability? An attempt at an answer to your first question anyway. The electrons surrounding an atom need to obey energy level (and other) rules. As you mention distance, if you imagine that the further away the ...


1

Electron as a standing wave Yes, the electron is a standing wave. See atomic orbitals on Wikipedia: "The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves". I couldn't understand how come Bohr who interpreted electron as a particle, formulated an equation for electron's angular ...


1

Yes, to the notion that it is really interactions between charged particles with quantum natures. The "walls" of the slit are really the outer electrons of the surface atoms of the substance that are either covalently or metallically bonded to each other. Those bound electrons form a potential surface that is not exactly square, but when viewed from the ...


1

Just extending Anna's answer. In order to observe the interference pattern which your probing electron(s) make, its associated de Broglie wavelength has to be several tens of nanometers at least. The wavelength of the electrons consisting the wall is approximately the distance between the atoms of the wall which is at least a few orders of magnitude lower ...


1

In quantum mechanics we use a wave function to describe the quantum state that an electron is in. It basically tells you where you are likely to find a particle. Notice how it is typically introduced as a function of position. That is because by taking the value at a certain location and squaring it we get the probability of finding it there.


1

Basically any measurement is on wave function |ψ⟩ is done by operator X such that X|ψ⟩ results observable x with some probability. is not entirely correct. In quantum mechanics a system is described by a state $|\psi\rangle\in\mathcal{H}$ and a set of self-adjoint operators $A_1,\ldots,A_n$ representing the observables that we want to measure. Moreover, ...


1

I was also puzzled with the way Zettili defined Δx and Δk to derive uncertainty relations. His definition is not the same as the well known FWHM=2.3548σ. First we need to realize that |ψ(x)|^2 and |ϕ(k)|^2 have the standard Gaussian form (1/σ√2π) Exp[−(1/2) (x/σ)^2] with a^2=2σ^2 --not |ψ(x)| and |ϕ(k)|. He defines the half maxima (0.6 to be precise) of ...


1

I) Let us here phrase the problem in the context of some position operator $\hat{q}$ of QM for simplicity. The generalization to QFT can formally be achieved by replacing the position operator $\hat{q}$ with a quantum field $\hat{\psi}({\bf x})$. We know that the overlap with Minkowski (M) signature is given as a path integral ...


1

Did you cover the uncertainty principle? In quantum mechanics there is and uncertainty between energy and time: $$ \Delta E \Delta t > \frac{h}{4\pi}$$ this means that if you try to measure Energy with perfect accuracy you will have a great uncertainly in time (actually an infinity uncertainty). I guess this is what the professor was referring to, and ...


1

Strangely enough, when you do a quantum mechanics experiment, you get a result that says something about what you already know. If you place a detector at one (or both) of the slits, you watch individual particles go through the slits, and you get a particle result from your detector (e.g., no interference pattern). If you don't know which slit the ...



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