How to Derive the Continuity Equation for a Particle in a 1D Potential?

In summary, Homework Equations:-The Schrodinger equation can be used to calculate the probability density for a particle in a one dimensional potential.-The current density is given by j(x,t)=-\frac{i\hbar}{2m}(\psi^{*}\frac{\partial^{2}{\psi}}{\partial{x^2}}+\frac{\partial{\psi^{*}}}{\partial{x}}\frac{\partial{\psi}}{\partial{x}}-\psi\frac{\partial^{2}{\psi^{*}}}{\partial{x^2}}+\frac{\partial{\psi}}{\partial{x}}
  • #1
whatisreality
290
1

Homework Statement


There's a particle moving in a 1D potential V(x) with mass m. The particle's normalised wavefunction is ψ(x,t). Use the time dependent Schrodinger equation to show that ##\frac{\partial{\rho}}{\partial{t}} + \frac{\partial{j}}{\partial{x}} = 0##
Where
##j(x,t) = -\frac{i\hbar}{2m}(\psi^{*} \frac{\partial{\psi}}{\partial{x}} - \psi \frac{\partial{\psi}}{\partial{x}})##

I also have to show that j(x,t) is real. All I know about j is that it has to be equal to the magnitude of ψ(x,t)##^{2}##.

Homework Equations


Time dependent Schrodinger equation:
##i\hbar \frac{\partial{\psi}}{\partial{t}} = -\frac{\hbar^{2}}{2m} \frac{\partial^{2}{\psi}}{\partial{x^2}} + V(x) \psi##

The probability density ρ = |ψ(x,t)|^2 or ψψ*.

The Attempt at a Solution


I'm having a bit of trouble with the calculus element. Pretty sure I'm differentiating wrong AND integrating wrong, but anyway, here's what I got:
First I thought I'd calculate ##\frac{\partial{j}}{\partial{x}}## since j is given. I got
##-\frac{i\hbar}{2m}(\psi^{*}\frac{\partial^{2}{\psi}}{\partial{x^2}}+\frac{\partial{\psi}}{\partial{x^2}}\psi^{*}-\psi\frac{\partial^{2}{\psi^{*}}}{\partial{x^2}}+\frac{\partial{\psi}}{\partial{x}}\frac{\partial{\psi}}{\partial{x}})##.

Then I thought I would solve the Schrodinger equation. But (and I know this isn't exactly a good reason) the question is only worth five marks! Which makes me think that I might not have to solve the Schrodinger equation. And I wouldn't actually know how to solve it anyway...

So is my first calculation right? And do I need to solve the Schrodinger equation in order to answer this question? If I do... how do I do it??
 
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  • #2
whatisreality said:
Then I thought I would solve the Schrodinger equation.
Why do you think solving the Schrodinger equation would be useful? Besides, it is not possible to solve the Schrodinger equation in closed form for general [itex]V(x)[/itex]

There is a mistake in your expression for the current density. There should be a complex conjugation on the second term: [tex]j(x,t) \equiv -\frac{i\hbar}{2m} \left(\psi^{*} \partial_{x} \psi - \psi \partial_{x} \psi^{*} \right)[/tex]
Calculating [itex]\partial_{x} j(x,t)[/itex] is a good place to start - do correct the complex conjugation mistakes though.
Have you tried calculating [itex]\partial_{t} \rho[/itex]?
 
  • #3
Fightfish said:
Why do you think solving the Schrodinger equation would be useful? Besides, it is not possible to solve the Schrodinger equation in closed form for general [itex]V(x)[/itex]

There is a mistake in your expression for the current density. There should be a complex conjugation on the second term: [tex]j(x,t) \equiv -\frac{i\hbar}{2m} \left(\psi^{*} \partial_{x} \psi - \psi \partial_{x} \psi^{*} \right)[/tex]
Calculating [itex]\partial_{x} j(x,t)[/itex] is a good place to start - do correct the complex conjugation mistakes though.
Have you tried calculating [itex]\partial_{t} \rho[/itex]?
So it represents current density! Interesting. OK, I'll sort out the conjugation mistakes - if I add the *, should it be correctly differentiated though?

I have tried calculating ρ for a start, but pretty unsuccessfully. I think I broke some rules along the way, while trying to solve the schrodinger equation, the potential was still in the equation at the end. I'll post where I got to...
 
  • #4
So corrected version for partial derivative of j: ##-\frac{i\hbar}{2m}(\psi^{*}\frac{\partial^{2}{\psi}}{\partial{x^2}}+\frac{\partial{\psi^{*}}}{\partial{x}}\frac{\partial{\psi}}{\partial{x}}-\psi\frac{\partial^{2}{\psi^{*}}}{\partial{x^2}}-\frac{\partial{\psi}}{\partial{x}}\frac{\partial{\psi^{*}}}{\partial{x}})##
 
  • #5
Fightfish said:
Why do you think solving the Schrodinger equation would be useful? Besides, it is not possible to solve the Schrodinger equation in closed form for general [itex]V(x)[/itex]

There is a mistake in your expression for the current density. There should be a complex conjugation on the second term: [tex]j(x,t) \equiv -\frac{i\hbar}{2m} \left(\psi^{*} \partial_{x} \psi - \psi \partial_{x} \psi^{*} \right)[/tex]
Calculating [itex]\partial_{x} j(x,t)[/itex] is a good place to start - do correct the complex conjugation mistakes though.
Have you tried calculating [itex]\partial_{t} \rho[/itex]?
As for ##\rho##, actually I know my method was wrong. I made ∂Ψ/∂t the subject and then integrated. I don't know how to solve the Schrodinger equation, so I can't find ##\rho##.
 
  • #6
You don't need to solve the Schrodinger equation to "find" [itex]\rho[/itex].
[tex]\partial_{t} (\psi^{*}\psi) = \psi \partial_{t} \psi^{*} + \psi^{*} \partial_{t} \psi[/tex]
Now, use the Schrodinger equation to replace [itex]\partial_{t} \psi^{*} [/itex] and [itex]\partial_{t} \psi[/itex]
 
  • #7
Fightfish said:
You don't need to solve the Schrodinger equation to "find" [itex]\rho[/itex].
[tex]\partial_{t} (\psi^{*}\psi) = \psi \partial_{t} \psi^{*} + \psi^{*} \partial_{t} \psi[/tex]
Now, use the Schrodinger equation to replace [itex]\partial_{t} \psi^{*} [/itex] and [itex]\partial_{t} \psi[/itex]
Got it, thank you!
 

FAQ: How to Derive the Continuity Equation for a Particle in a 1D Potential?

1. What is a particle in a 1D potential V(x) system?

A particle in a 1D potential V(x) system is a theoretical model used in physics to study the behavior of a single particle moving in a one-dimensional space under the influence of a potential energy function V(x).

2. What is the significance of studying a particle in a 1D potential V(x) system?

Studying a particle in a 1D potential V(x) system allows us to understand the behavior of particles in a simplified environment and can provide insights into more complex systems. It also has practical applications in fields such as quantum mechanics and materials science.

3. How is the particle's motion affected by the potential energy function in a 1D potential V(x) system?

The potential energy function V(x) determines the forces acting on the particle, which in turn affects its motion. The particle will move towards regions of lower potential energy and experience forces towards regions of higher potential energy.

4. Can the particle's energy change in a 1D potential V(x) system?

Yes, the particle's energy can change in a 1D potential V(x) system. This change can occur when the particle interacts with the potential energy function or when external forces act on the particle.

5. What types of potential energy functions are commonly used in a 1D potential V(x) system?

Some commonly used potential energy functions in a 1D potential V(x) system include the harmonic oscillator potential, the square well potential, and the barrier potential. These functions can be used to model a variety of physical systems and phenomena.

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