Benutzer:Viberion/ Spielwiese

First, let's start with the definition of a complex plane wave. A complex plane wave is a type of mathematical function that describes a wave-like behavior in two-dimensional space. It is defined as:

$\psi (x,y)=A\exp \left(i\left(kx-\omega t\right)\right)$

where A is the amplitude of the wave, k is the wavenumber, $\omega$ is the angular frequency, and t is time.

Next, let's consider the energy of the electron in the wave. The energy of a particle is typically given by the equation $E={\frac {p^{2}}{2m}}$, where p is the momentum of the particle and m is its mass. In the case of the electron, we can also include the electric potential energy of the electron in the electric field, which is given by the equation $V=-e\phi$, where e is the charge of the electron and $\phi$ is the electric potential.

Therefore, the total energy of the electron in the wave can be written as:

$E={\frac {p^{2}}{2m}}-e\phi$

Now, we can use the principles of quantum mechanics to relate the momentum and position of the electron to the wave function $\psi (x,y)$. According to the principles of quantum mechanics, the position and momentum of a particle are represented by operators, which are denoted by ${\hat {x}}$ and ${\hat {p}}$, respectively. These operators act on the wave function to produce a new wave function that represents the position or momentum of the particle.

The position and momentum operators are related by the equation:

${\hat {p}}=-i\hbar \nabla$

where $\hbar$ is the reduced Planck constant and $\nabla$ is the gradient operator.

Using this relationship, we can substitute the momentum operator into the equation for the energy of the electron to obtain:

$E=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}-e\phi$

Finally, we can use the time-dependent Schrödinger equation to describe how the wave function of the electron changes over time. The time-dependent Schrödinger equation is given by:

$i\hbar {\frac {\partial \psi }{\partial t}}={\hat {H}}\psi$

where ${\hat {H}}$ is the Hamiltonian operator, which represents the total energy of the system.

Substituting the expression for the energy of the electron into the time-dependent Schrödinger equation, we obtain the Schrödinger equation for the electron in an electric potential:

$i\hbar {\frac {\partial \psi }{\partial t}}=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi -e\phi \psi$

This is the Schrödinger equation for the electron in an electric potential. I hope this helps! Let me know if you have any other questions.

Quantum tunneling is a phenomenon that occurs when a particle, such as an electron, passes through a potential barrier that it does not have enough energy to surmount. This phenomenon is described by the Schrödinger equation, which is a fundamental equation in quantum mechanics that describes how the wave function of a physical system changes over time.

To derive quantum tunneling, we can start by considering a particle with wave function $\psi(x)$ moving in one dimension and encountering a potential barrier of height $V_0$ and width $a$. The potential barrier is described by the equation:

$V(x)={\begin{cases}0&{\text{if }}x<0{\text{ or }}x>a\ V_{0}&{\text{if }}0\leq x\leq a\end{cases}}$

We can use the Schrödinger equation to describe how the wave function of the particle changes as it encounters the potential barrier. The Schrödinger equation is given by:

$i\hbar {\frac {\partial \psi }{\partial t}}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}\psi }{\partial x^{2}}}+V(x)\psi$

To solve this equation, we need to consider two different regions: the region before the potential barrier, where x < 0, and the region after the potential barrier, where x > a. In each of these regions, we can assume that the wave function takes the form:

$\psi (x)={\begin{cases}A_{1}\exp \left(i{\frac {k_{1}}{\hbar }}x\right)&{\text{if }}x<0\ A_{2}\exp \left(i{\frac {k_{2}}{\hbar }}x\right)&{\text{if }}x>a\end{cases}}$

where $A_{1}andA_{2}$ are the amplitudes of the wave function in each region, and $k_{1}andk_{2}$ are the wave numbers.

Now, we can use the Schrödinger equation to find the wave function in the region between the potential barrier, where $0\leq x\leq a$. In this region, the potential is non-zero, so the Schrödinger equation becomes:

$i\hbar {\frac {\partial \psi }{\partial t}}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}\psi }{\partial x^{2}}}+V_{0}\psi$

To solve this equation, we can assume that the wave function in this region is a linear combination of the form:

$\psi (x)=C_{1}\exp \left(i{\frac {k_{1}}{\hbar }}x\right)+C_{2}\exp \left(i{\frac {k_{2}}{\hbar }}x\right)$

where $C_{1}$ and $C_{2}$ are constants to be determined. Substituting this expression into the Schrödinger equation and solving for $C_{1}$ and $C_{2}$, we obtain:

$C_{1}=A_{1}\exp \left(i{\frac {k_{1}}{\hbar }}x\right)$

$C_{2}=A_{2}\exp \left(i{\frac {k_{2}}{\hbar }}x\right)$

To continue, we need to find the wave function in the region between the potential barrier, where $0 \le x \le a$. We can do this by substituting the expressions for $C_1$ and $C_2$ into the wave function in this region, which gives:

$\psi (x)=A_{1}\exp \left(i{\frac {k_{1}}{\hbar }}x\right)+A_{2}\exp \left(i{\frac {k_{2}}{\hbar }}x\right)$

This is the wave function of the particle in the region between the potential barrier. Now, we can use this wave function to find the probability of the particle tunneling through the potential barrier.

To do this, we need to consider the wave function at the edges of the potential barrier, where x = 0 and x = a. At these points, the wave function must be continuous, which means that the amplitudes of the wave function on either side of the barrier must be equal. Therefore, we can set $A_{1}=A_{2}=A$ and solve for $k_{1}$ and $k_{2}$ in terms of A and the energy E of the particle.

To do this, we can use the Schrödinger equation to write the energy of the particle in terms of the wave function and the potential. The Schrödinger equation is given by:

$i\hbar {\frac {\partial \psi }{\partial t}}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}\psi }{\partial x^{2}}}+V(x)\psi$

If we multiply both sides of this equation by $\psi$, where $\psi$ is the complex conjugate of the wave function, we obtain:

$i\hbar {\frac {\partial \psi ^{*}\psi }{\partial t}}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}\psi ^{*}\psi }{\partial x^{2}}}+V(x)\psi ^{*}\psi$

If we integrate both sides of this equation over all space, we obtain:

$i\hbar \int {\frac {\partial \psi ^{*}\psi }{\partial t}},dx=-{\frac {\hbar ^{2}}{2m}}\int {\frac {\partial ^{2}\psi ^{*}\psi }{\partial x^{2}}},dx+\int V(x)\psi ^{*}\psi ,dx$

The left-hand side of this equation is the rate of change of the probability of the particle being in a particular location, which is zero. Therefore, the right-hand side must also be zero, which gives us:

$-{\frac {\hbar ^{2}}{2m}}\int {\frac {\partial ^{2}\psi ^{*}\psi }{\partial x^{2}}},dx+\int V(x)\psi ^{*}\psi ,dx=0$

To continue, we need to use the expressions for the wave function and the potential to find the energy of the particle in terms of the amplitude $A$ and the wave numbers $k_{1}$ and $k_{2}$.

To do this, we can substitute the expressions for the wave function and the potential into the Schrödinger equation, which gives:

$i\hbar {\frac {\partial \left(A^{2}\right)}{\partial t}}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}\left(A^{2}\right)}{\partial x^{2}}}+V_{0}A^{2}$

If we integrate both sides of this equation over all space, we obtain:

$i\hbar \int {\frac {\partial \left(A^{2}\right)}{\partial t}},dx=-{\frac {\hbar ^{2}}{2m}}\int {\frac {\partial ^{2}\left(A^{2}\right)}{\partial x^{2}}},dx+\int V_{0}A^{2},dx$

The left-hand side of this equation is the rate of change of the probability of the particle being in a particular location, which is zero. Therefore, the right-hand side must also be zero, which gives us:

$-{\frac {\hbar ^{2}}{2m}}\int {\frac {\partial ^{2}\left(A^{2}\right)}{\partial x^{2}}},dx+\int V_{0}A^{2},dx=0$

If we substitute the expressions for the wave function and the potential into this equation, we obtain:

$-{\frac {\hbar ^{2}}{2m}}\int {\frac {\partial ^{2}\left(A^{2}\right)}{\partial x^{2}}},dx+V_{0}\int A^{2},dx=0$

We can simplify this equation by noting that the first term on the left-hand side is equal to the second term on the right-hand side, which gives us:

$\int {\frac {\partial ^{2}\left(A^{2}\right)}{\partial x^{2}}},dx=V_{0}\int A^{2},dx$

This equation tells us that the rate of change of the probability of the particle being in a particular location is equal to the potential energy of the particle.

Now, we can use this equation to find the energy of the particle in terms of A, $k_{1}$, and $k_{2}$. To do this, we can substitute the expression for the wave function into the left-hand side of the equation and evaluate the integral. This gives us:

${\frac {\partial ^{2}\left(A^{2}\right)}{\partial x^{2}}}=V_{0}A^{2}$

If we substitute the equation for the wave function, we get

${\frac {\partial ^{2}\left(A^{2}\right)}{\partial x^{2}}}=A^{2}\left({\frac {k_{1}^{2}}{\hbar ^{2}}}+{\frac {k_{2}^{2}}{\hbar ^{2}}}\right)$

If we evaluate this equation at the edges of the potential barrier, where x = 0 and x = a, we obtain:

${\frac {\partial ^{2}\left(A^{2}\right)}{\partial x^{2}}}{\Bigg |}_{x=0}=A^{2}\left({\frac {k_{1}^{2}}{\hbar ^{2}}}+{\frac {k_{2}^{2}}{\hbar ^{2}}}\right)=V_{0}A^{2}$

${\frac {\partial ^{2}\left(A^{2}\right)}{\partial x^{2}}}{\Bigg |}_{x=a}=A^{2}\left({\frac {k_{1}^{2}}{\hbar ^{2}}}+{\frac {k_{2}^{2}}{\hbar ^{2}}}\right)=V_{0}A^{2}$

These equations tell us that the rate of change of the probability of the particle being in a particular location is equal to the potential energy of the particle at the edges of the potential barrier.

Now, we can use these equations to find the energy of the particle in terms of A, $k_{1}$, and $k_{2}$. To do this, we can solve the equations for $k_{1}^{2}$ and $k_{2}^{2}$ and substitute the resulting expressions into the energy equation, which is given by:

$E={\frac {p^{2}}{2m}}={\frac {\hbar ^{2}}{2m}}\left(k_{1}^{2}+k_{2}^{2}\right)$

Solving the equations for $k_{1}^{2}$ and $k_{2}^{2}$, we obtain:

$k_{1}^{2}={\frac {2mV_{0}}{\hbar ^{2}}}A^{2}-{\frac {k_{2}^{2}}{\hbar ^{2}}}$

$k_{2}^{2}={\frac {2mV_{0}}{\hbar ^{2}}}A^{2}-{\frac {k_{1}^{2}}{\hbar ^{2}}}$

Substituting these expressions into the energy equation, we obtain:

$E={\frac {\hbar ^{2}}{2m}}\left({\frac {2mV_{0}}{\hbar ^{2}}}A^{2}-{\frac {k_{2}^{2}}{\hbar ^{2}}}+{\frac {2mV_{0}}{\hbar ^{2}}}A^{2}-{\frac {k_{1}^{2}}{\hbar ^{2}}}\right)$

This equation gives us the energy of the particle in terms of the amplitude A

The Boltzmann transport equation is a mathematical equation that describes the evolution of the distribution function of a system of particles over time. It is named after the Austrian physicist Ludwig Boltzmann, who made significant contributions to the development of statistical mechanics.

To derive the Boltzmann transport equation from Liouville's theorem, we first need to recall the statement of Liouville's theorem:

${\frac {df}{dt}}=0$

where $f(\mathbf {q} ,\mathbf {p} ,t)$ is the distribution function of the system, which describes the probability density of finding a particle at a particular position $\mathbf {q}$ and momentum $\mathbf {p}$ at time t. The distribution function is related to the density in phase space $\rho (\mathbf {q} ,\mathbf {p} ,t)$ by:

$\rho (\mathbf {q} ,\mathbf {p} ,t)={\frac {N}{V}}f(\mathbf {q} ,\mathbf {p} ,t)$

where N is the total number of particles in the system and V is the volume of the system.

To derive the Boltzmann transport equation, we need to consider the effect of collisions on the distribution function. When two particles collide, their positions and momenta may change, which will affect the distribution function.

To take into account the effect of collisions, we can write the Boltzmann transport equation as:

${\frac {df}{dt}}={\frac {1}{N}}\int \int d^{3}q_{2}d^{3}p_{2},\omega (\mathbf {q} _{1},\mathbf {p} _{1};\mathbf {q} _{2},\mathbf {p} _{2})\left[f_{2}f_{1}-f_{1}'f_{2}'\right]$

where $f_{1}andf_{2}$ are the distribution functions of the colliding particles before the collision, and f_1' and f_2' are the distribution functions after the collision. The function $\omega (\mathbf {q} _{1},\mathbf {p} _{1};\mathbf {q} _{2},\mathbf {p} _{2})$ is the collision frequency, which gives the probability per unit time that a collision will occur between particles 1 and 2 at the given positions and momenta.

This equation states that the change in the distribution function of a particle over time is equal to the difference between the distribution function after a collision and the distribution function before the collision, multiplied by the collision frequency. This takes into account the effect of collisions on the distribution function, and allows us to describe the evolution of the distribution function over time.

This equation is known as the Boltzmann transport equation, and it is a key equation in the study of statistical mechanics. It is used to describe the behavior of systems of particles in equilibrium and out of equilibrium, and has applications in a wide range of fields, including physics, engineering, and chemistry.

To derive the usual form of the Boltzmann transport equation from the equation given above, we can start by expressing the distribution function in terms of the density in phase space:

$f(\mathbf {q} ,\mathbf {p} ,t)={\frac {V}{N}}\rho (\mathbf {q} ,\mathbf {p} ,t)$

Substituting this expression into the equation I gave in my first response, we get:

${\frac {df}{dt}}={\frac {1}{N}}\int \int d^{3}q_{2}d^{3}p_{2},\omega (\mathbf {q} _{1},\mathbf {p} _{1};\mathbf {q} _{2},\mathbf {p} _{2})\left[f_{2}f_{1}-f_{1}'f_{2}'\right]$

${\frac {d}{dt}}\left({\frac {V}{N}}\rho \right)={\frac {1}{N}}\int \int d^{3}q_{2}d^{3}p_{2},\omega (\mathbf {q} _{1},\mathbf {p} _{1};\mathbf {q} _{2},\mathbf {p} _{2})\left[f_{2}f_{1}-f_{1}'f_{2}'\right]$

Then, we can use the chain rule to rewrite the time derivative on the left-hand side as:

${\frac {\partial \rho }{\partial t}}+\mathbf {v} \cdot \nabla _{\mathbf {q} }\rho +\nabla _{\mathbf {p} }V\cdot \nabla _{\mathbf {p} }\rho ={\frac {1}{N}}\int \int d^{3}q_{2}d^{3}p_{2},\omega (\mathbf {q} _{1},\mathbf {p} _{1};\mathbf {q} _{2},\mathbf {p} _{2})\left[f_{2}f_{1}-f_{1}'f_{2}'\right]$

where $\mathbf {v} =\mathbf {p} /m$ is the velocity of a particle, where $m$ is its mass, and $V(\mathbf {q} )$ is the potential energy of the particle at position $\mathbf {q}$ . The gradient operators $\nabla _{\mathbf {q} }$ and $\nabla _{\mathbf {p} }<math>aredefinedas:<math>\nabla _{\mathbf {q} }=\left({\frac {\partial }{\partial q_{1}}},{\frac {\partial }{\partial q_{2}}},{\frac {\partial }{\partial q_{3}}}\right)$

$\nabla _{\mathbf {p} }=\left({\frac {\partial }{\partial p_{1}}},{\frac {\partial }{\partial p_{2}}},{\frac {\partial }{\partial p_{3}}}\right)$

Finally, we can express the density in phase space in terms of the distribution function using the expression we derived earlier: ${\frac {\partial f}{\partial t}}+\mathbf {v} \cdot \nabla _{\mathbf {q} }f-\nabla _{\mathbf {p} }V\cdot \nabla _{\mathbf {p} }f={\frac {1}{N}}\int \int d^{3}q_{2}d^{3}p_{2},\omega (\mathbf {q} _{1},\mathbf {p} _{1};\mathbf {q} _{2},\mathbf {p} _{2})\left[f_{2}f_{1}-f_{1}'f_{2}'\right]$