Relation Between the Wave Function and Probability Density: Difference between revisions

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{{Quantum Mechanics A}}
{{Quantum Mechanics A}}
The quantity <math>|\Psi(\textbf{r},t)|^2</math> can be interpreted as probability density. In order for this to be true, two conditions must be met. First, the probability amplitude must be positive semi-definite (equal to or greater than zero). This condition is trivial because <math>|\Psi(\textbf{r},t)|^2 \!</math> is always a positive function. Second, the probability density, integrated over all space, must be equal to one:
The quantity <math>|\Psi(\textbf{r},t)|^2</math> can be interpreted as probability density. In order for us to do so, two conditions must be met. First, the probability amplitude must be positive semi-definite (equal to or greater than zero). This condition is trivial because <math>|\Psi(\textbf{r},t)|^2 \!</math> is always a non-negative function. Second, the probability density, integrated over all space, must be equal to one:


:<math>\int_{-\infty}^{\infty}d^3\textbf{r}\,|\Psi(\textbf{r},t_0)|^2=1\;\;\forall t</math>
:<math>\int_{-\infty}^{\infty}d^3\textbf{r}\,|\Psi(\textbf{r},t)|^2=1</math>


We will show that, if this relation is satisfied for a specific time, then it is satisfied for all times shortly. Because of the fact that we may interpret <math>|\Psi(\textbf{r},t)|^2</math> as a probability density, we may calculate expectation values of observables, such as position and momentum, in terms of it.  These expectation values are
We will show that, if this relation is satisfied for a specific time, then it is satisfied for all times shortly.
 
Because of the fact that we may interpret <math>|\Psi(\textbf{r},t)|^2</math> as a probability density, we may calculate expectation values of observables, such as position and momentum, in terms of it.  In general, the expectation value of an observable <math>Q(\textbf{r},\textbf{p};t)</math> is given by
 
:<math>\langle Q(\textbf{r},\textbf{p};t)\rangle=\int_{-\infty}^{\infty}d^3\textbf{r}\,\Psi^{\ast}(\textbf{r},t)Q\left (\textbf{r},-i\hbar\nabla;t\right )\Psi(\textbf{r},t).</math>
 
In particular, the expectation value of a position coordinate <math>x_i</math> is


:<math>\langle x_i\rangle=\int_{-\infty}^{\infty}d^3\textbf{r}\,x_i|\Psi(\textbf{r},t)|^2</math>
:<math>\langle x_i\rangle=\int_{-\infty}^{\infty}d^3\textbf{r}\,x_i|\Psi(\textbf{r},t)|^2</math>


and
and that for a component of momentum <math>p_i</math> is


:<math>\langle p_i\rangle=\int_{-\infty}^{\infty}d^3\textbf{r}\,\Psi^{\ast}(\textbf{r},t)\left (-i\hbar\frac{\partial}{\partial x_i}\right )\Psi(\textbf{r},t).</math>
:<math>\langle p_i\rangle=\int_{-\infty}^{\infty}d^3\textbf{r}\,\Psi^{\ast}(\textbf{r},t)\left (-i\hbar\frac{\partial}{\partial x_i}\right )\Psi(\textbf{r},t).</math>
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We will now show that the solution to the [[Schrödinger equation]] conserves probability, i.e. the probability to find the particle somewhere in the space does not change with time.  To see that it does, consider  
We will now show that the solution to the [[Schrödinger equation]] conserves probability, i.e. the probability to find the particle somewhere in the space does not change with time.  To see that it does, consider  


:<math>i\hbar\frac{\partial}{\partial t}\psi(\textbf{r},t)=\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right]\psi(\textbf{r},t)</math>  
<math>i\hbar\frac{\partial}{\partial t}\Psi(\textbf{r},t)=\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right]\Psi(\textbf{r},t)</math>
 
Now multiply both sides by the complex conjugate of <math>\psi(\textbf{r},t):\!</math>


Now multiply both sides by the complex conjugate of <math>\psi(\textbf{r},t) \!</math>:
<math>i\hbar\Psi^{\ast}(\textbf{r},t)\frac{\partial}{\partial t}\Psi(\textbf{r},t)=\Psi^{\ast}(\textbf{r},t)\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right]\Psi(\textbf{r},t)</math>


:<math>i\hbar\psi^*(\textbf{r},t)\frac{\partial}{\partial t}\psi(\textbf{r},t)=\psi^*(\textbf{r},t)\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right]\psi(\textbf{r},t)</math> <br/>
Now, take the complex conjugate of this entire expression:
Now, take the complex conjugate of this entire expression: <br/>
:<math>-i\hbar\psi(\textbf{r},t)\frac{\partial}{\partial t}\psi^*(\textbf{r},t)=\psi(\textbf{r},t)\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right]\psi^*(\textbf{r},t)</math> <br/>
and taking the difference of the above equations, we finally find


:<math>\frac{\partial}{\partial t} \left( \psi^*(\textbf{r},t)\psi(\textbf{r},t)\right) + \frac{\hbar}{2im} \nabla \cdot \left[\psi^*(\textbf{r},t)\nabla \psi(\textbf{r},t)-(\nabla\psi^*(\textbf{r},t)) \psi(\textbf{r},t)\right]=0</math>
<math>-i\hbar\Psi(\textbf{r},t)\frac{\partial}{\partial t}\Psi^{\ast}(\textbf{r},t)=\Psi(\textbf{r},t)\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right]\Psi^{\ast}(\textbf{r},t)</math>
 
Taking the difference of the above equations, we finally find
 
<math>\frac{\partial}{\partial t} \left( \Psi^{\ast}(\textbf{r},t)\Psi(\textbf{r},t)\right) + \frac{\hbar}{2im} \nabla \cdot \left[\Psi^{\ast}(\textbf{r},t)\nabla \Psi(\textbf{r},t)-(\nabla\Psi^{\ast}(\textbf{r},t)) \Psi(\textbf{r},t)\right]=0</math>
   
   
Note that this is in the form of a continuity equation,


Note that this is in the form of a continuity equation
<math>\frac{\partial}{\partial t} \rho(\textbf{r},t) + \nabla \cdot \textbf{j}(\textbf{r},t)=0,</math>
 
:<math>\frac{\partial}{\partial t} \rho(\textbf{r},t) + \nabla \cdot \textbf{j}(\textbf{r},t)=0</math>
   
   
where  
where  


:<math>\rho(\textbf{r},t)=\psi^*(\textbf{r},t) \psi(\textbf{r},t)\!</math>  
<math>\rho(\textbf{r},t)=\Psi^{\ast}(\textbf{r},t)\Psi(\textbf{r},t)\!</math>  


is the probability density, and   
is the probability density, and   


:<math>\textbf{j}(\textbf{r},t)=\frac{\hbar}{2im}\left[\psi^*(\textbf{r},t)\nabla \psi(\textbf{r},t)-(\nabla\psi^*(\textbf{r},t)) \psi(\textbf{r},t)\right]</math>  
:<math>\textbf{j}(\textbf{r},t)=-\frac{i\hbar}{2m}\left[\Psi^{\ast}(\textbf{r},t)\nabla \Psi(\textbf{r},t)-(\nabla\Psi^{\ast}(\textbf{r},t)) \Psi(\textbf{r},t)\right]</math>  


is the probability current.
is the probability current.


Once we know that the densities and currents constructed from the solution of the Schrödinger equation satisfy the continuity equation, it is easy to show that the probability is conserved.  
Once we know that the densities and currents constructed from the solution of the Schrödinger equation satisfy the continuity equation, it is easy to show that the probability is conserved.
To see that note:
 
To see this, note that
 
:<math>\frac{d}{dt}\int d^3\textbf{r} |\Psi(\textbf{r},t)|^2=-\int d^3\textbf{r}(\nabla\cdot \textbf{j})=-\oint d\textbf{A} \cdot \textbf{j} =0.</math>
 
Here, we used the fact that the wave function is assumed to vanish outside of the boundary, and thus the current vanishes as well.  Therefore, we see that the normalization of the wave function does not change over time, so that we only need to normalize it at one instant in time, as asserted earlier.
 
==Problems==
 
'''(1)''' Consider a particle moving in a potential field <math>V(\textbf{r}).</math>
 
'''(a)''' Prove that the average energy is <math>\langle E\rangle=\int W\,d^3\textbf{r}=\int\left (\frac{\hbar^2}{2m}\nabla\Psi^{\ast}\cdot\nabla\Psi+\Psi^{\ast}V\Psi\right )\,d^3\textbf{r},</math> where <math>W</math> is energy density.
 
'''(b)''' Prove the energy conservation equation, <math>\frac{\partial W}{\partial t}+\nabla \cdot \textbf{S}=0,</math> where <math>\textbf{S}=-\frac{\hbar^2}{2m}\left (\frac{\partial\Psi^{\ast}}{\partial t}\nabla\Psi + \frac{\partial\Psi}{\partial t}\nabla\Psi^{\ast}\right )</math> is the energy flux density.
 
[[Phy5645/Energy_conservation|Solution]]
 
 
'''(2)''' Assume that the Hamiltonian for a system of <math>N</math> particles is <math>\hat{H}=-\sum_{i=1}^{N}\frac{\hbar}{2m}\nabla_{i}^{2}+\sum_{i=1}^{N}\rho_{ij}(|\textbf{r}_{i}-\textbf{r}_{j}|)</math>, and <math>\Psi(\textbf{r}_{1},\textbf{r}_{2},\ldots,\textbf{r}_{N};t)</math> is the wave fuction.  Defining
 
<math>\rho(\textbf{r},t)=\sum\rho_{i}(\textbf{r},t),</math>
 
<math>\textbf{j}(\textbf{r},t)=\sum\textbf{j}_{i}(\textbf{r},t),</math>
 
<math>\rho_{1}(\textbf{r}_{1},t)=\int\cdots\int d^{3}\textbf{r}_{2}\,d^{3}\textbf{r}_{3}\,\cdots\,d^{3}\textbf{r}_{N}\,\Psi^{\star}\Psi,</math>
 
<math>\textbf{j}_{1}(\textbf{r}_{1},t)=-\frac{i\hbar}{2m}\int\cdots\int d^{3}\textbf{r}_{2}\,d^{3}\textbf{r}_{3}\,\cdots\,d^{3}\textbf{r}_{N}\,(\Psi^{\star}\nabla_{1}\Psi-\Psi\nabla_{1}\Psi^{\star}),</math>
 
and similarly for the other <math>\rho_i</math> and <math>\textbf{j}_i</math>, prove the following relation:


:<math>\frac{\partial}{\partial t}\int d^3r |\psi(\textbf{r},t)|^2=-\int d^3r(\nabla\cdot \textbf{j})=-\oint d\textbf{A} \cdot \textbf{j} =0</math>  
<math>\frac{\partial\rho}{\partial t}+\nabla\cdot\textbf{j}=0</math>


where we used the divergence theorem which relates the volume integrals to surface integrals of a vector field.  Since the wavefunction is assumed to vanish outside of the boundary, the current vanishes as well. So, this proved the probability of finding the particle in the whole space is independent of time.
[[Phy5645/schrodingerequationhomework2|Solution]]

Latest revision as of 13:37, 8 August 2013

Quantum Mechanics A
SchrodEq.png
Schrödinger Equation
The most fundamental equation of quantum mechanics; given a Hamiltonian Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}} , it describes how a state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi\rangle} evolves in time.
Basic Concepts and Theory of Motion
UV Catastrophe (Black-Body Radiation)
Photoelectric Effect
Stability of Matter
Double Slit Experiment
Stern-Gerlach Experiment
The Principle of Complementarity
The Correspondence Principle
The Philosophy of Quantum Theory
Brief Derivation of Schrödinger Equation
Relation Between the Wave Function and Probability Density
Stationary States
Heisenberg Uncertainty Principle
Some Consequences of the Uncertainty Principle
Linear Vector Spaces and Operators
Commutation Relations and Simultaneous Eigenvalues
The Schrödinger Equation in Dirac Notation
Transformations of Operators and Symmetry
Time Evolution of Expectation Values and Ehrenfest's Theorem
One-Dimensional Bound States
Oscillation Theorem
The Dirac Delta Function Potential
Scattering States, Transmission and Reflection
Motion in a Periodic Potential
Summary of One-Dimensional Systems
Harmonic Oscillator Spectrum and Eigenstates
Analytical Method for Solving the Simple Harmonic Oscillator
Coherent States
Charged Particles in an Electromagnetic Field
WKB Approximation
The Heisenberg Picture: Equations of Motion for Operators
The Interaction Picture
The Virial Theorem
Commutation Relations
Angular Momentum as a Generator of Rotations in 3D
Spherical Coordinates
Eigenvalue Quantization
Orbital Angular Momentum Eigenfunctions
General Formalism
Free Particle in Spherical Coordinates
Spherical Well
Isotropic Harmonic Oscillator
Hydrogen Atom
WKB in Spherical Coordinates
Feynman Path Integrals
The Free-Particle Propagator
Propagator for the Harmonic Oscillator
Differential Cross Section and the Green's Function Formulation of Scattering
Central Potential Scattering and Phase Shifts
Coulomb Potential Scattering

The quantity Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi(\textbf{r},t)|^2} can be interpreted as probability density. In order for us to do so, two conditions must be met. First, the probability amplitude must be positive semi-definite (equal to or greater than zero). This condition is trivial because Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi(\textbf{r},t)|^2 \!} is always a non-negative function. Second, the probability density, integrated over all space, must be equal to one:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{-\infty}^{\infty}d^3\textbf{r}\,|\Psi(\textbf{r},t)|^2=1}

We will show that, if this relation is satisfied for a specific time, then it is satisfied for all times shortly.

Because of the fact that we may interpret Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi(\textbf{r},t)|^2} as a probability density, we may calculate expectation values of observables, such as position and momentum, in terms of it. In general, the expectation value of an observable Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q(\textbf{r},\textbf{p};t)} is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle Q(\textbf{r},\textbf{p};t)\rangle=\int_{-\infty}^{\infty}d^3\textbf{r}\,\Psi^{\ast}(\textbf{r},t)Q\left (\textbf{r},-i\hbar\nabla;t\right )\Psi(\textbf{r},t).}

In particular, the expectation value of a position coordinate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_i} is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle x_i\rangle=\int_{-\infty}^{\infty}d^3\textbf{r}\,x_i|\Psi(\textbf{r},t)|^2}

and that for a component of momentum Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_i} is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle p_i\rangle=\int_{-\infty}^{\infty}d^3\textbf{r}\,\Psi^{\ast}(\textbf{r},t)\left (-i\hbar\frac{\partial}{\partial x_i}\right )\Psi(\textbf{r},t).}

Conservation of Probability

We will now show that the solution to the Schrödinger equation conserves probability, i.e. the probability to find the particle somewhere in the space does not change with time. To see that it does, consider

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\hbar\frac{\partial}{\partial t}\Psi(\textbf{r},t)=\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right]\Psi(\textbf{r},t)}

Now multiply both sides by the complex conjugate of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(\textbf{r},t):\!}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\hbar\Psi^{\ast}(\textbf{r},t)\frac{\partial}{\partial t}\Psi(\textbf{r},t)=\Psi^{\ast}(\textbf{r},t)\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right]\Psi(\textbf{r},t)}

Now, take the complex conjugate of this entire expression:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -i\hbar\Psi(\textbf{r},t)\frac{\partial}{\partial t}\Psi^{\ast}(\textbf{r},t)=\Psi(\textbf{r},t)\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right]\Psi^{\ast}(\textbf{r},t)}

Taking the difference of the above equations, we finally find

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial t} \left( \Psi^{\ast}(\textbf{r},t)\Psi(\textbf{r},t)\right) + \frac{\hbar}{2im} \nabla \cdot \left[\Psi^{\ast}(\textbf{r},t)\nabla \Psi(\textbf{r},t)-(\nabla\Psi^{\ast}(\textbf{r},t)) \Psi(\textbf{r},t)\right]=0}

Note that this is in the form of a continuity equation,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial t} \rho(\textbf{r},t) + \nabla \cdot \textbf{j}(\textbf{r},t)=0,}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho(\textbf{r},t)=\Psi^{\ast}(\textbf{r},t)\Psi(\textbf{r},t)\!}

is the probability density, and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \textbf{j}(\textbf{r},t)=-\frac{i\hbar}{2m}\left[\Psi^{\ast}(\textbf{r},t)\nabla \Psi(\textbf{r},t)-(\nabla\Psi^{\ast}(\textbf{r},t)) \Psi(\textbf{r},t)\right]}

is the probability current.

Once we know that the densities and currents constructed from the solution of the Schrödinger equation satisfy the continuity equation, it is easy to show that the probability is conserved.

To see this, note that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dt}\int d^3\textbf{r} |\Psi(\textbf{r},t)|^2=-\int d^3\textbf{r}(\nabla\cdot \textbf{j})=-\oint d\textbf{A} \cdot \textbf{j} =0.}

Here, we used the fact that the wave function is assumed to vanish outside of the boundary, and thus the current vanishes as well. Therefore, we see that the normalization of the wave function does not change over time, so that we only need to normalize it at one instant in time, as asserted earlier.

Problems

(1) Consider a particle moving in a potential field Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(\textbf{r}).}

(a) Prove that the average energy is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle E\rangle=\int W\,d^3\textbf{r}=\int\left (\frac{\hbar^2}{2m}\nabla\Psi^{\ast}\cdot\nabla\Psi+\Psi^{\ast}V\Psi\right )\,d^3\textbf{r},} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W} is energy density.

(b) Prove the energy conservation equation, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial W}{\partial t}+\nabla \cdot \textbf{S}=0,} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \textbf{S}=-\frac{\hbar^2}{2m}\left (\frac{\partial\Psi^{\ast}}{\partial t}\nabla\Psi + \frac{\partial\Psi}{\partial t}\nabla\Psi^{\ast}\right )} is the energy flux density.

Solution


(2) Assume that the Hamiltonian for a system of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N} particles is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{H}=-\sum_{i=1}^{N}\frac{\hbar}{2m}\nabla_{i}^{2}+\sum_{i=1}^{N}\rho_{ij}(|\textbf{r}_{i}-\textbf{r}_{j}|)} , and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi(\textbf{r}_{1},\textbf{r}_{2},\ldots,\textbf{r}_{N};t)} is the wave fuction. Defining

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho(\textbf{r},t)=\sum\rho_{i}(\textbf{r},t),}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \textbf{j}(\textbf{r},t)=\sum\textbf{j}_{i}(\textbf{r},t),}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho_{1}(\textbf{r}_{1},t)=\int\cdots\int d^{3}\textbf{r}_{2}\,d^{3}\textbf{r}_{3}\,\cdots\,d^{3}\textbf{r}_{N}\,\Psi^{\star}\Psi,}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \textbf{j}_{1}(\textbf{r}_{1},t)=-\frac{i\hbar}{2m}\int\cdots\int d^{3}\textbf{r}_{2}\,d^{3}\textbf{r}_{3}\,\cdots\,d^{3}\textbf{r}_{N}\,(\Psi^{\star}\nabla_{1}\Psi-\Psi\nabla_{1}\Psi^{\star}),}

and similarly for the other Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho_i} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \textbf{j}_i} , prove the following relation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial\rho}{\partial t}+\nabla\cdot\textbf{j}=0}

Solution

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