I am unsure as to the nature of the potential

In summary: Remember, the SE is a differential equation - you get a solution by solving for the constants that make it true for all x. Then, if it is true for all x, it is normalized and is a valid wavefuncton. So, solve for C, and then for ##\alpha## in terms of the constants.In summary, a particle with mass m and electric charge e is confined to move in one dimension along the x-axis and experiences a potential of infinity for x<0 and a Coulomb potential of -e^2/4πε0x for x≥0. By substituting the given wave function, u(x)=Cxexp(-αx), into the Schrödinger equation
  • #1
Darrenm95
7
0

Homework Statement


A particle with mass m and electric charge e is confined to move in one dimension along the x -axis. It experiences the following potential:
##V(x) = {\infty}## for ## x{\lt0}##
##V(x) = -e^2/4\pi\epsilon_0x## for ## x \geq 0 ##
(Note: the way the question is written down features no parentheses around ##4\pi\epsilon_0x##
a) Describe the potential experienced by the particle.
b) (This question just asks to write down the TISE which I can do).
c)For the region ## x \geq 0 ##, by substituting in the Schrödinger equation, show that the wave function
## u(x)= Cxexp(\text{-}\alpha x) ##
can be a satisfactory solution of the Schrodinger equation so long as the constant ##\alpha## is suitably chosen. Determine the unique expression for ##\alpha## in terms of m , e and other fundamental constants. Note that C is a normalisation constant.

Homework Equations


TISE

The Attempt at a Solution


For a) I am unsure to what the potential actually is and for c) I took the second derivative of u(x) and substituted it into the Schrodinger equation but it was more of a stab in the dark then anything else. I also thought about using boundary conditions to find an equation for ##\alpha## but given that my only boundary condition is 0, I've not managed to figure out a way of finding anything meaningful from doing this.
 
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  • #2
Based on the offered eigenfunction in part (c), I think we can safely conclude that the the potential is [itex]V(x) = \frac{e^{2}}{4\pi\epsilon x}[/itex] for [itex]x\geq 0 [/itex]. What does this remind you of?

Yes, "taking the second derivative of u(x) and substituting it into the Schrodinger equation" is the right approach. Can you show your work on this?
 
  • #3
There should be a negative sign in the potential, otherwise ##\alpha## will be negative and the wavefunction won't be normalizable.
 
  • #4
Darrenm95 said:

Homework Statement


A particle with mass m and electric charge e is confined to move in one dimension along the x -axis. It experiences the following potential:
##V(x) = {\infty}## for ## x{\lt0}##
##V(x) = e^2/4\pi\epsilon_0x## for ## x \geq 0 ##
(Note: the way the question is written down features no parentheses around ##4\pi\epsilon_0x##
a) Describe the potential experienced by the particle.
b) (This question just asks to write down the TISE which I can do).
c)For the region ## x \geq 0 ##, by substituting in the Schrödinger equation, show that the wave function
## u(x)= Cxexp(\text{-}\alpha x) ##
can be a satisfactory solution of the Schrodinger equation so long as the constant ##\alpha## is suitably chosen. Determine the unique expression for ##\alpha## in terms of m , e and other fundamental constants. Note that C is a normalisation constant.

Homework Equations


TISE

The Attempt at a Solution


For a) I am unsure to what the potential actually is and for c) I took the second derivative of u(x) and substituted it into the Schrodinger equation but it was more of a stab in the dark then anything else. I also thought about using boundary conditions to find an equation for ##\alpha## but given that my only boundary condition is 0, I've not managed to figure out a way of finding anything meaningful from doing this.

Note on proper LaTeX use: never write ##exp(-\alpha x)##: it looks ugly and can lead to ambiguities. Write, instead, ##\exp(-\alpha x)##, which looks much better and is unambiguous. You do that by writing "\exp" instead of "exp". The same holds for sin, cos, tan, ln, log, max. min, lim, etc.---all the standard functions. You get ##\sin##, ##\cos##, ##\tan##, ##\ln##, ##\log##, ##\max##, ##\min##, ##\lim## instead of ##sin##, ##cos##, ##tan##, ##ln##, ##log##, ##max##, ##min##, ##lim##, etc.

Also, you do not need to say "exp(\text{-} \alpha x)"; LaTeX is smart enough and powerful enough to know how to parse the much simpler "\exp(-\alpha x)". The "\text{-}" plays no role here, so is a waste of your time and resources.
 
  • #5
Ray Vickson said:
Note on proper LaTeX use: never write ##exp(-\alpha x)##: it looks ugly and can lead to ambiguities. Write, instead, ##\exp(-\alpha x)##, which looks much better and is unambiguous. You do that by writing "\exp" instead of "exp". The same holds for sin, cos, tan, ln, log, max. min, lim, etc.---all the standard functions. You get ##\sin##, ##\cos##, ##\tan##, ##\ln##, ##\log##, ##\max##, ##\min##, ##\lim## instead of ##sin##, ##cos##, ##tan##, ##ln##, ##log##, ##max##, ##min##, ##lim##, etc.

Also, you do not need to say "exp(\text{-} \alpha x)"; LaTeX is smart enough and powerful enough to know how to parse the much simpler "\exp(-\alpha x)". The "\text{-}" plays no role here, so is a waste of your time and resources.
Thank you for the advice, I've never used LaTex before and I used the text on the minus sign because I preferred the shorter length.
 
  • #6
blue_leaf77 said:
There should be a negative sign in the potential, otherwise ##\alpha## will be negative and the wavefunction won't be normalizable.
Yes, sorry, I missed the minus sign
 
  • #7
Darrenm95 said:
For a) I am unsure to what the potential actually is and for c) I took the second derivative of u(x) and substituted it into the Schrodinger equation but it was more of a stab in the dark then anything else. I also thought about using boundary conditions to find an equation for ##\alpha## but given that my only boundary condition is 0, I've not managed to figure out a way of finding anything meaningful from doing this.
(a) I concur with the others here - the potential is the coulomb potential - with a hard surface at x=0, or you can think of x as a radial distance.
Later in your education you will be solving the SE with this sort of thing as a radiual component - right now your prof seems to want you to have some experience with the math so it's quite artificial.

(c) You are on to a good start - after substituting, just use ordinary algebra to solve for alpha.
 

What is the meaning of "potential" in science?

The term "potential" in science refers to the possibility or likelihood of something happening or being developed in the future. It can also refer to the capacity or ability of an object or system to perform a certain function.

How do scientists determine the potential of a substance or system?

Scientists use various methods such as experiments, observations, and mathematical models to determine the potential of a substance or system. These methods involve collecting data and analyzing it to make predictions about the behavior or properties of the substance or system.

What factors affect the nature of potential in a scientific context?

The nature of potential in a scientific context can be affected by various factors such as the properties of the substance or system, external influences, and the conditions under which it is being studied. Other factors may include the presence of other substances or forces, temperature, and pressure.

How can understanding the nature of potential be useful in scientific research?

Understanding the nature of potential can be useful in scientific research as it allows scientists to make accurate predictions and develop theories about the behavior of a substance or system. It also helps in identifying potential applications and limitations of a substance or system, which can aid in further research and development.

Can the potential of a substance or system change over time?

Yes, the potential of a substance or system can change over time due to various factors such as changes in external conditions, interactions with other substances or forces, and the introduction of new variables. This is why ongoing research and experimentation are important in understanding and predicting the potential of a substance or system.

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