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5

The square of the wavefunction, $|\Psi(x,t)|^2$, is the probability density function for finding the particle. This means that the probability of finding the particle in an interval of (infinitesimal) width $\mathrm dx$ at position $x$ equals $|\Psi(x,t)|^2\mathrm dx$. On occasion, however, authors will drop the $\mathrm dx$ if it is convenient and does not ...


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

The Schrodinger equation is an approximation because it ignores relativistic effects and it ignores spin. However, aside from these limits it applies to any system and not just electrons. The trouble is that the Schrodinger equation, and indeed most partial differential equations, are impossible to solve except in a few special cases. Since we have no ...


3

Given the wavefunction $\psi(x)$, the probability to find any particle within an interval $[x_0,x_0+\Delta x]$ is $$ P([x_0,x+\Delta x_0]) = \int_{x_0}^{x_0+\Delta x} \lvert\psi(x)\rvert^2\mathrm{d}x$$ i.e. $\lvert\psi(x)\rvert^2$ is not a probability, but a probability density that has to be integrated over a set of non-zero measure to yield a notion of ...


3

To understand what these orbitals are, you first have to understand the notion of superposition in quantum mechanics. In regular classical physics, a particle or a system must be in a definite state. A car is at a particular mile marker on a highway, moving at a particular speed. The Moon orbits around the Earth with a particular velocity at a particular ...


3

The visualization method you choose is directly and completely determined by the information you need to see regarding your state. For the states of a single bosonic mode, there are multiple different visualization methods, and they all have their pros and cons. In particular, there is a direct trade-off between the amount of information you can display on a ...


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

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

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}) = ...


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

The probability of just one of your atoms getting through the wall is very low, multiply that probability (which is way lower than 1) by the probabilty that all of your atoms go through, you get a number so small that the universe would almost certainly have ended long before you got through the wall. A 70 Kg human body has approximately 7 X $10^{27}$ ...


2

The standard procedure is the following: starting from $ \langle nlm|\,\partial^2_z\,| n'l'm'\rangle $ insert the identity operator with respect to the position basis $$ 1 = \int d\textbf{r} |\textbf{r}\rangle\otimes\langle \textbf{r}| $$ to have $$ \int d\textbf{r}\, \langle nlm\, |\,\partial^2_z\,|\textbf{r}\rangle\cdot\langle \textbf{r}|n'l'm'\rangle. ...


2

I don't want to be too much precise, but the Schrödinger equation ($i \dot{\psi}= H\psi$ to avoid confusion) has at most an unique solution under very general assumptions on the Hamiltonian operator $H$, even if you see it as a liner equation in the more general setting of Banach spaces. In particular, it is not a priori necessary that $H$ is self-adjoint ...


2

First, realize we are doing an approximation when we are evaluating the coefficient $A_p$ in the Fourier series $$ \psi(x) = \sum_p A_p \psi_p(x)$$ by the integral $$ A_p = \int_{-\infty}^\infty \psi_p^*(x)\psi(x)\mathrm{d}x$$ since the limits should really be the boundary of the interval on which the Fourier series is periodic. Furthermore, $\psi(x) = ...


2

So, I suppose that $Φ(k)$ is the probability density of the momentum. Is this true? Almost. $\Phi(k)$ is the probability amplitude for the momentum of the particle. The probability density is obtained as usual by squaring the amplitude, giving $|\Phi(k)|^2$. For a free particle, all values of momentum are always allowed, which enables the superposition ...


2

In the case of an infinite superposition of eigenstates it becomes more complicated but we can still write a general expression for it. If $$\psi(x) = \sum_{n=0}^\infty a_n \phi_n(x)$$ where the $\phi_n$ are the eigenstates of the Hamiltonian. The time dependent wavefunction will look like: $$\Psi(x,t) = \sum_{n=0}^\infty a_n \phi_n(x) T_n(t)$$ where $T_n = ...


2

In general, the integral $$ V := \int \mathrm{d} \mu = \int 1 \mathrm{d}\mu$$ is the integration of the identity over the space the measure $\mu$ is defined on, and should be intuitively understood as the volume of the space with respect to the measure. (This is usually only finite for compact spaces.) Normalizing the measure means sending $\mu \mapsto ...


1

Experimental physicists are very visual people. They are more interested in seeing the wavefunction. Calculating it is the work of a theoretical physicist. "Catching sight of the elusive wavefunction", would be a good read for you at this point. As you proceed with Griffiths, you will see that in experiments we measure certain quantities, called the ...


1

The wave function is connected to an experimentally observable value through its complex conjugate square, which gives the probability of an interaction or a decay happening; from this a crossection for the interaction can be predicted, number of events versus some variable in appropriate units. Example: the experiment can measure very many decays of the ...


1

If I understand the question. Assume you have a quantum oscillator, then if the system is an eigenvector of the hamiltonian, then a series of measurements will give a series of results reflecting eigenvalues of the system. The appearance of these eigenvalues must obey at a number of many measurements the probabilities of each eigenvalue to appear. It's ...


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Integration over pure quantum states usually refers to the Haar measure, i.e., the unitarily invariant measure. Vaguely speaking, you assign the same volume (=weight in the integral) to any two set of states which are related by an arbitrary unitary rotation $U$. In the case of one qubit, this is equivalent to integrating over the Bloch sphere; i.e., we ...


1

You are making confusion between two different things: the first is why the wave function of two identical particles is what it is, the second is the probability of an event happening in several ways (which has nothing to do with physics but is just the definition of unions of probabilities). As per the first question: given $\mathcal{H}_1$ and ...


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Solving time-dependent SE as danielsmw mentioned above good starting point. $$ i\hbar\frac{\partial\psi}{\partial t}=H\psi $$ $$ \frac{d\psi}{\psi}=\frac{H}{i\hbar}dt $$ $$ log(\psi)\mid^{\psi}_{\psi_{0}}=-\frac{iHt}{\hbar}\mid^{t}_{t_0} $$ suppose $\psi=|\alpha,t>$ and $\psi_{0}=|\alpha, t_{0}>$ $$ \psi=e^{-\frac{iH(t-t_{0})}{\hbar}}\psi_0 $$ Under ...


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, ...



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