Deerive Hte Continuity Equation Using Schrodinger Equation

Consider a "free particle state" corresponding to $V=0$ : a particle traveling in space with no potential holding it down. Since microscopic particles behave as waves in quantum mechanics, a good model for the particle is a "wavepacket" formed by a superposition of plane waves at different momenta $\hbar k$ , corresponding to the wavefunction:

$\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{\phi(k)}{e}^{i(kx-\omega t) }\,dk.$

Notably, since the form of $\phi (k)$ is not given, the above statement is just that $\Psi (x,t)$ is the inverse Fourier transform of some function $\phi (k)$ .

Using de Broglie's equations of momentum $p=\hbar k=\frac{h}{\nu}$ and energy $E= \hbar \omega = \frac{{p}^{2}}{2m}$ , the equation for the wave-packet equation can be rewritten as

$\Psi(x,t) = \frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}{\phi(p)}{e}^{i(px-Et)/ \hbar}\, dp.$

To attempt to write down an equation of motion for $\Psi (x,t)$ , it is instructive to look at the different derivatives of $\Psi (x,t)$ that are possible. Physicists typically believe the equation of state of an arbitrary system should be given in terms of first or second derivatives at the most, so that the initial configuration and motion of a system are enough to specify the state at all future times.

Taking the time derivative and two derivatives in space yields

$\begin{aligned} \frac { \partial \Psi }{ \partial t } &=\frac { 1 }{ \sqrt { 2\pi \hbar } } \int _{ -\infty }^{ \infty }{ \phi (p){ \left( \frac { -iE }{ \hbar } \right) e }^{ i(px-Et)/\hbar }\,dp }\\ \frac { \partial \Psi }{ \partial x } &=\frac { 1 }{ \sqrt { 2\pi \hbar } } \int _{ -\infty }^{ \infty }{ \phi (p){ \left( \frac { ip }{ \hbar } \right) e }^{ i(px-Et)/\hbar }\,dp } \\ \frac { { \partial }^{ 2 }\Psi }{ \partial { x }^{ 2 } } &=\frac { 1 }{ \sqrt { 2\pi \hbar } } \int _{ -\infty }^{ \infty }{ \phi (p){ \left( \frac { -{ p }^{ 2 } }{ { \hbar }^{ 2 } } \right) e }^{ i(px-Et)/\hbar }\,dp }. \end{aligned}$

Multiplying the time derivative by $i\hbar$ and the second spatial derivative by $\frac{-{\hbar}^{2}}{2m}$ and using the de Broglie relations, one has the following interesting equality:

$i\hbar \frac { \partial \Psi }{ \partial t } = \frac{-{\hbar}^{2}}{2m}\frac { { \partial }^{ 2 }\Psi }{ \partial { x }^{ 2 } }.$

The right-hand size has units of energy; it is the kinetic energy of the quantum-mechanical state (a statement which is further justified below). To generalize this free-particle derivation to a particle in an arbitrary potential, one needs only to add $V\Psi$ to the right hand side. This is sufficient to obtain the result, the time-dependent Schrödinger equation:

$i\hbar \frac { \partial \Psi }{ \partial t } = \frac{-{\hbar}^{2}}{2m}\frac { { \partial }^{ 2 }\Psi }{ \partial { x }^{ 2 } }+ V\Psi.$

Quantum mechanics is inherently linear, which means linear algebra is the language of QM. Thus, it is most appropriate to write the Schrödinger equation in operator form. The mathematics of operators in QM is further discussed in the wavefunctions and measurement wiki.

For now, note that it is useful to define the energy operator $\hat{E}$ and momentum operator $\hat{p}$ :

$\hat{E} = i\hbar \partial_t, \qquad \hat{p} = -i\hbar \partial_x.$

This discussion is motivated by the description of a particle as a superposition of plane wave states. Considering a plane wave $\psi = e^{\frac{i}{\hbar} (px - Et)}$ , applying these operators yields

$\begin{aligned} \hat{E} \psi &= i\hbar \partial_t e^{\frac{i}{\hbar} (px - Et)} = i \hbar (-\frac{i}{\hbar} E) \psi = E \psi \\ \hat{p} \psi &= -i\hbar \partial_x e^{\frac{i}{\hbar} (px - Et)} = -i \hbar (\frac{i}{\hbar} p) \psi = p \psi . \end{aligned}$

Therefore, the eigenvalues of these operators correspond to the energy and momentum of a wavefunction, respectively.

Using this fact, it is straightforward to formulate the Schrödinger equation directly from operators. Note that the energy of a classical particle is

$E = \frac{p^2}{2m} + V,$

where $V$ is the potential energy. Considering the system in quantum mechanics requires "promoting" all variables to operators acting on a wavefunction. The energy becomes the Hamiltonian operator $\hat{H}$ :

$\hat{H} \psi = \frac{\hat{p}^2}{2m} \psi + V \psi.$

Equating the Hamiltonian operator $\hat{H}$ with the energy operator $\hat{E}$ , one obtains the time-dependent Schrödinger equation again from the operator formulation:

$i\hbar \partial_t \psi = -\frac{\hbar^2}{2m } \partial_x^2 \psi + V\psi.$

Recalling the form of the energy operator $\hat{E} = i\hbar \partial_t$ , the equation can also be rewritten in the extremely compact form:

$\hat{E} \psi = \hat{H} \psi,$

where $\hat{E}$ generates time evolution and $\hat{H}$ depends on the kinetic and potential energies of the quantum particle.

Find the energy of the state given at some time by

$\psi(x) = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} e^{-\frac{m\omega x^2}{2\hbar}}$

in the harmonic oscillator potential, $V(x) = \frac12 m \omega^2 x^2$ , given that this state is an eigenfunction of the corresponding Hamiltonian.

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Since the state is an eigenfunction of the given Hamiltonian, one only needs to apply the Hamiltonian operator to this state:

$\begin{aligned} \hat{H} \psi &= -\frac{\hbar^2}{2m} \partial_x^2 \psi + \frac12 m\omega^2 x^2 \psi \\ &= -\frac{\hbar^2}{2m} \partial_x \left(-\frac{m\omega x}{\hbar} \psi\right) + \frac12 m\omega^2 x^2 \psi \\ &=-\frac{\hbar^2}{2m} \left( \left(-\frac{m\omega x}{\hbar}\right)^2 \psi -\frac{m\omega }{\hbar} \psi \right) + \frac12 m\omega^2 x^2 \psi \\ &= -\frac12 m \omega^2 x^2 \psi + \frac12 \hbar \omega \psi + \frac12 m\omega^2 x^2 \psi \\ &= \frac12 \hbar \omega \psi. \end{aligned}$

Reading off the eigenvalue, the energy is $E = \frac12 \hbar \omega$ . This is the ground state of the quantum harmonic oscillator.

$\frac12 \hbar \omega$ $2 \hbar \omega$ $\frac32 \hbar \omega$ $\hbar \omega$

Find the energy of the state given at some time by

$\psi(x) = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} \sqrt{\frac{2m\omega}{\hbar}} xe^{-\frac{m\omega x^2}{2\hbar}}$

in the harmonic oscillator potential $V = \frac12 m \omega^2 x^2$ .

A last formulation of the time-dependent Schrödinger equation is the method from unitary evolution. Recall that a unitary operator is one that preserves the norm of a state. To this end, suppose that states evolve in time unitarily so that the total probability of a wavefunction is always one:

$\psi (x,t) = \hat{U} (t) \psi (x,0),$

with $\hat{U}^{\dagger} \hat{U} = \hat{1}$ at all times.

Plugging this state into the Schrödinger equation, one can verify that it is a solution at all times:

$i\hbar \frac{\partial \hat{U} \psi (x,0)}{\partial t} = \hat{H} \hat{U} \psi(x,0) \implies i\hbar \frac{\partial \hat{U} }{\partial t} = \hat{H} \hat{U},$

where the second equation is purely an equation for the operator $\hat{U} (t)$ . Assuming the Hamiltonian is time-independent, the solution to this first-order ODE is

$\hat{U} = e^{-i\hat{H} t / \hbar}.$

Therefore, the state

$\psi(x,t) = e^{-i\hat{H} t / \hbar} \psi(x,0)$

obeys the Schrödinger equation at all times, and it has been shown that unitary evolution is equivalent to Schrödinger evolution.

Since any state $\psi(x,0)$ can be written in a superposition of energy eigenstates as

$\psi(x,0) = \sum_n \phi_n (x),$

the general solution to the Schrödinger equation given an initial state can be found by individually evolving each energy eigenstate:

$\psi(x,t) = \sum_n e^{-iE_n t / \hbar} \phi_n (x).$

A particle in quantum mechanics has the non-normalized wavefunction at time $t=0$ :

$\psi(x) = 2 \phi_2 + i \phi_5,$ where the $\phi_n$ are the orthonormal eigenstates of some Hamiltonian with energies $E_n = n^3$ .

Find the normalized wavefunction at all times.

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First, normalize the wavefuntion at time $t=0$ . Since Schrödinger evolution is unitary, the state will stay normalized at all times if it starts normalized. The normalized wavefunction is $\psi (x) = \frac{2}{\sqrt{5}} \phi_2 + \frac{i}{\sqrt{5}} \phi_5.$ To check this, compute the square of $\psi (x)$ and use the orthonormality of the $\phi_n$ : $\left|\frac{2}{\sqrt{5}}\right|^2 + \left|\frac{i}{\sqrt{5}}\right|^2 = \frac{4+1}{5} = 1.$ Now, since the $\phi_n$ are stationary states, each can be time-evolved individually, and the answer is $\psi(x,t) = \frac{2}{\sqrt{5}} e^{-8it/\hbar} \phi_2 (x) + \frac{i}{\sqrt{5}} e^{-125 it/\hbar} \phi_5 (x).$

$-i e^{-it/\hbar} \phi_1 + e^{-it/(3\hbar)} \phi_3$ $-\frac{i}{\sqrt{2}} \phi_1 + \frac{1}{\sqrt{2}} \phi_3$ $-\frac{i}{\sqrt{2}} e^{-it/\hbar} \phi_1 + \frac{1}{\sqrt{2}} e^{-3it/\hbar} \phi_3$ $-\frac{i}{\sqrt{2}} e^{-it/\hbar} \phi_1 + \frac{1}{\sqrt{2}} e^{-it/(3\hbar)} \phi_3$

A particle in quantum mechanics has the non-normalized wavefunction at $t=0$ :

$\psi(x) = -i \phi_1 + \phi_3,$

where the $\phi_n$ are the orthonormal eigenstates of some Hamiltonian with energies $E_n = \frac{1}{n}$ .

Find the normalized wavefunction $\psi(x,t)$ at all times.

One can rewrite the Schrödinger equation in a way that avoids the time derivative by considering states of definite energy. For such states, $\hat{E} \psi = E\psi$ by definition. The Schrodinger equation then becomes $\hat{H} \psi = E\psi$ , where $E$ is now a constant and not an operator. Written out fully, the time-independent Schrödinger equation is thus

$E \psi = -\frac{\hbar^2}{2m} \partial_x^2 \psi + V \psi.$

These eigenstates $\psi$ of the energy operator are typically called stationary states or determinate states. From the compact form $\hat{H} \psi = E \psi$ , one can see that the time-independent Schrödinger equation is an eigenvector equation for $\hat{H}$ . The set of values of $E$ for which there exists a corresponding $\psi$ satisfying this equation is called the spectrum of the Hamiltonian $H$ . If two or more linearly independent eigenfunctions share the same eigenvalue, we call that eigenvalue (and the spectrum containing that eigenvalue) degenerate. Degeneracies play an important role in describing the atom in much of quantum chemistry and solid-state physics; for instance, the orbitals of the electron in the hydrogen atom are degenerate states.

Degeneracy of the electronic states of the hydrogen atom. At the $n$ th energy level, there are $2\sum_{\ell = 0}^{n-1} (2\ell + 1) = 2n^2$ states of equal energy, including the degeneracies in the $m$ and $s$ quantum numbers. In the above diagram, all states of equal $\ell$ are combined into a single line [2].

A classic basic model for a particle confined in some spatial region is the one-dimensional "particle in a box" potential, which is described by a free particle potential surrounded by infinitely tall walls:

Potential energy function for the 1D "particle in a box" setup [3].

Find the stationary states and corresponding energies for the "particle in a box" potential:

$V(x) = \begin{cases} 0 , \qquad &0< x < L \\ \infty, \qquad &\text{ otherwise}. \end{cases}$

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Since the potential is infinite outside $0<x<L$ , there is zero probability of finding the particle in those regions and correspondingly the wavefunction must be zero there. In between the walls, the time-independent Schrödinger equation reads

$E \psi = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \psi.$

The eigenfunctions of the Hamiltonian are the one-dimensional eigenfunctions of the Laplacian:

$\phi_n (x) = A \sin (kx),$

where $A$ is some normalization constant. Because $\phi_n (0) = 0$ , there is no cosine term above. Enforcing the boundary condition $\phi_n (L) = 0$ requires

$kL = \pi n \implies k_n = \frac{\pi n}{L}$

for any positive integer $n$ . The eigenfunctions that are consistent with the boundary conditions of the potential are thus

$\phi_n (x) = A\sin (k_n x),$

with the corresponding eigenvalues/energies (from the time-independent Schrödinger equation)

$E_n = \frac{\hbar^2 k_n^2}{2m} = \frac{\hbar^2 \pi^2 n^2}{mL^2}.$

Normalizing the eigenfunctions, one finds

$\phi_n (x) = \sqrt{\frac{2}{L}} \sin (k_n x).$

A: classical motion of a particle in a 1D box. B-D: the time evolution of energy eigenfunctions in the stationary states of the one-dimensional "particle in a box" potential. E-F: time evolution of states which are not energy eigenfunctions [1].

The set of real numbers The set of perfect squares The set of non-negative integers The set of non-negative real numbers

What is the spectrum of the free-particle Hamiltonian (redefined up to a constant)

$\hat{H} = - \frac{d^2}{dx^2}\:?$

There is no degeneracy in the spectrum. Every eigenvalue but one has a degeneracy of order 2. Every eigenvalue has a degeneracy of order 2. Exactly one eigenvalue has a degeneracy, which is of order 2.

A particle constrained to move on a ring can be described by one parameter, the angle $\phi$ on the ring with respect to some coordinates. In quantum mechanics, an otherwise free particle constrained to move on this ring has the Hamiltonian

$\hat{H} = -\frac{d^2}{d\phi^2}.$

Which of the following correctly describes the spectrum of this Hamiltonian?

[1] Image from https://en.wikipedia.org/wiki/Particle_in_a_box under Creative Commons licensing for reuse and modification.

[2] Image from https://upload.wikimedia.org/wikipedia/commons/6/64/Hydrogen_energy_levels.png under Creative Commons licensing for reuse and modification.

[3] Image from https://en.wikipedia.org/wiki/Quantum_mechanics under Creative Commons licensing for reuse and modification.

[4] Griffiths, David J. Introduction to Quantum Mechanics. Second Edition. Pearson: Upper Saddle River, NJ, 2006.

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Source: https://brilliant.org/wiki/schrodinger-equation/

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