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12

The total probability of all mutually excluding alternatives must always be 100%, so it is conserved. The conservation law in the spacetime tend to be "local", so just like for the charge conservation, we may derive the conservation of the probability in Schrödinger's equation from the local continuity equation $$ \frac{\partial \rho}{\partial t} + \mathbf ...


6

There is a fundamental theorem already conjectured by von Neumann but proved just at the end of 1900 by Solèr (in addition to a partial result already obtained by by Piron in the sixties) which establishes (relying on the theory of orthomodular lattices and projective geometry) that the general phenomenology of Quantum Mechanics can be described only by ...


5

First, a note about the Hamiltonian and its time derivatives. I think that it is misleading to write that the Hamiltonian $$ H = i\hbar\frac{d}{dt}, $$ although the time-dependent Schrodinger equation is of course $$ H\psi = i\hbar\frac{d}{dt} \psi. $$ To evaluate e.g. $\frac{d}{dt}H$ you should consider $H=H(p, q, t)$, rather than $H = i\hbar\frac{d}{dt}$. ...


4

A wavefunction can be nowhere continuous. It is enough that it belongs to $L^2(\mathbb R)$, so, in general, no regularity conditions are imposed on values attained at every given point of $\mathbb R$. It is only required that $\int_{\mathbb R} |\psi(x)|^2 dx < +\infty$. (Regularity conditions can be imposed when the wavefunction is required to belong to ...


4

You will not easily find the second one because in a sense it is trivial. Here is the single slit and double slit pattern from wikipedia . If you try to detect which slit the particle went through you get two single slits except for some experiments that are very careful not to disturb too much the wave functions of the setup. If you go to a lab that ...


4

To add to @Luboš Motl's great answer, I just want to mention the connection to the electric current and electric charge density from electrodynamics, which you may be more familiar with. Note that probability is unitless, so probability density has units of 1/Volume and probability current has units of 1/area*time. These are the same units as electric ...


3

From my limited knowledge of this subject, I would say that a non-normalizable wave-function wouldn't really make any physical sense. Remember, the wave-function is a function whose value squared evaluated between two points represents the probability that the particle will be found between those two points. So, the restriction that wave functions be ...


3

Let us consider the famous double-slit experiment with photons. With the usual set-up, we denote the number of photons passing through by $N$ and we will denote the number of photons which hit the film between $y$ and $y + \Delta y$ by $n(y)$. The probability that a photon will be detected between $y$ and $y+ \Delta y$ at a time $t$ is given by: ...


3

I think the answer is "because it works". Early in the development of QM, that interpretation was given to the wave function, and over the decades it has proven to be a useful interpretation. It works. Additionally, the fact that it is possible to define an associated current, and that there is a quantum mechanical expression that guarantees that ...


3

The general solution is the first one you gave, $$\psi(x,t) =Ae^{i(kx-\omega t)} + Be^{-i(kx+\omega t)}.$$ To get to the second one, you need to impose appropriate boundary conditions, which essentially means having no incoming particle flux from $+\infty$. Alternatively, you can see the solution $$\psi_k(x,t) =Ae^{i(kx-\omega t)}$$ as general enough to ...


2

This problem could be done more simply through the application of linear algebra. You want to prove that $$\langle \psi_1 - \psi_2 | \psi_1 + \psi_2 \rangle = 0$$ The inner product is analogous to the dot product of linear algebra, and it is distributive. Distributing, we find that $$\begin{aligned} \langle \psi_1 - \psi_2 | \psi_1 + \psi_2 \rangle &= ...


2

$|\Psi|^2=\Psi^*\Psi=(c_1^*\Psi_1^* + c_2^*\Psi_2^*)(c_1\Psi_1 + c_2\Psi_2)=c_1^*\Psi_2^*c_1\Psi_1+c_2^*\Psi^*c_2\Psi_2 + c_1^*\Psi_1^*c_2\Psi_2 + c_1\Psi_1c_2^*\Psi_2^* = |c_1\Psi_1|^2+|c_2\Psi_2|^2 + c_1\Psi_1c_2^*\Psi_2^* + (c_1\Psi_1c_2^*\Psi_2^*)^* = |c_1\Psi_1|^2+|c_2\Psi_2|^2 + 2Re(c_1\Psi_1c_2^*\Psi_2^*)$ Than substituting 2.9, you get 2.10.


2

From the Schroedinger equation, $$i\hbar\frac{\partial\psi}{\partial t}=\hat{H}\psi$$ simply divide by $i\hbar$: $$\frac{\partial\psi}{\partial t}=\frac{1}{i\hbar}\hat{H}\psi=-\frac{i}{\hbar}\hat{H}\psi\\$$ where we used $$ \frac{1}{i}\cdot\frac{i}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i $$ Now note that you've incorrectly applied the derivative. In quantum ...


1

If you suppose that $\psi$ is a real function, then as suggested by @Danu: $$\int_{-\infty}^\infty \dfrac{\partial^2{\psi^*}}{\partial x^2}\dfrac{\partial{\psi}}{\partial x}\, \rm{d}x = -\int_{-\infty}^\infty \dfrac{\partial{\psi^*}}{\partial x}\dfrac{\partial^2{\psi}}{\partial x^2}\, \rm{d}x,$$ and this gives zero, since $\psi=\psi^*$ and we always ...


1

Short answer The reason why a physical quantity such as probability is given by $|\Psi|^2$ rather than some other function of $\Psi$ is geometry, namely Pythagoras' theorem. If you have a vector which points from the origin to the $(\hat x,\hat y,\hat z)$ coordinates $(x,y,z)$, then the length $\ell$ is given by $\ell^2=x^2+y^2+z^2$. Why is this the ...


1

Note: ChocoPouce's answer is the same as mine but is more mathematical. You have a (spherically symmetric) probability density distribution $\rho$ in space (which we get from the square of the amplitude). The "radial probability density" is roughly the chance that the electron is at a given radius, say $r = 0.1\mathrm{nm}$? In other words, how much of this ...


1

The link uses a spherical shell element which is $4\pi r^2 \mathrm{d}r$ and has the dimension of a volume ($r^2$ is a surface and $\mathrm{d}r$ is a length. The wavefunction of the ground state is spherical, if it weren't the calculation should have been made using a spherical volume element such as $r^2\mathrm{d}r\mathrm{d}\theta\mathrm{d}\varphi$. There ...


1

The wavefunction must be either normalizable or the limit of a sequence of normalizable functions which in general are known as distributions (generalizations of functions). A well known example of a distribution is the Dirac delta "function," $\delta(x)$. If the spatial wavefunction is $\psi=\delta(x_0)$, then the momentum wavefunction will be of the form ...


1

The current density formula you consider is not right, because $\frac{d(\psi\psi^*)}{dt}$ is always 0. The current density is defined as $\bf{j}=(i\hbar/2m)(\psi\nabla\psi^*-\psi^*\nabla\psi)$. In your one dimensional case change $\nabla$ to $\partial/\partial x$, you will get your answer. A more physical definition of current density operator is this: ...


1

They have used the Schrödinger equation: $$i\hbar\frac{\partial \psi}{\partial t} = \hat{H} \psi$$ And hence: $$\frac{\partial \psi}{\partial t} = \frac{1}{i\hbar}\hat{H}\psi$$ And since $1/i = -i$: $$\frac{\partial \psi}{\partial t} = -\frac{i}{\hbar}\hat{H}\psi$$ Similarly for $\psi^*$ we take the complex conjugate: $$\frac{\partial \psi^*}{\partial t} = ...


1

To arrive at the 2nd term of the third line, they have simply employed the SE equation, $$i\hbar\frac{\partial \psi}{\partial t} = \hat{H}\psi$$ Re-arranging the above equation gives: $$\frac{\partial \psi}{\partial t} = -\frac{i}{\hbar}\hat{H}\psi$$ Therefore by replacing $\frac{\partial \psi}{\partial t}$ in the second line with ...


1

Part 1: As described in the text book, $\Psi$ is a linear combination of its eigenfunctions. $a_n$ is just a scalar multiple of that state, which basically tells you how much goes into that state, its the probability amplitude of that state. Therefore the probability to be in state $\Psi_n$ is equal to $\|a_n\|^2 = a_n a^*_n$. Part 2: It should be clear ...


1

One model is to say that the atom is in an impenetrable spherical box, and solve for the wavefunctions. See Y P Varshni Accurate wavefunctions for the confined hydrogen atom at high pressures J. Phys. B: At. Mol. Opt. Phys. 30 No 18 (28 September 1997) L589-L593. The Fermi Contact Term (electron density at the nucleus) greatly increases as the size of the ...



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