How Do You Calculate the Probability of Finding a Particle in the Ground State?

Are you trying to find the probability of finding the particle in the ground state or the energy of the ground state? In either case, you need to use the orthogonality of the eigenfunctions ##\psi_n(x)## and the coefficients ##c_n##. Can you try using this information to solve the problem?
  • #1
R-ckay
4
0
a particle is confined in a potential box 0≤x≤a. when t=0 wave function is given: ψ(x,0)=1/5(√6sin(∏x/a)+2sin(2∏x/a)). find probability of finding the particle in ground state and energy in ground state !

anyone please tell me how to solve the problem
 
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  • #2
Have you thought about the problem yourself before? In this forum you don't get the solution without showing some work, but help to find the solution yourself, and this helps you more than you might think!
 
  • #3
yes I have, sir
 
  • #4
Post your attempt at the solution.
 
  • #5
[itex]\int[/itex](boundary a to 0) |ψ(x,0)|^2=[ √6/√5sin(∏x/a)+2/√5sin(2∏x/a)][√6/√5sin(∏x/a)+2/√5sin(2∏x/a]
[itex]\int[/itex] [6/5(sin(∏x/a))^2+2√6/5sin(∏x/a)sin(∏x/a)+2√6/5sin(∏x/a)sin(2∏x/a)+4/5(sin(2∏x/a))^2

but I didn't find the final value. there's still "a" in the final result instead of a number of probability.
 
  • #6
R-ckay said:
[itex]\int[/itex](boundary a to 0) |ψ(x,0)|^2=[ √6/√5sin(∏x/a)+2/√5sin(2∏x/a)][√6/√5sin(∏x/a)+2/√5sin(2∏x/a]
[itex]\int[/itex] [6/5(sin(∏x/a))^2+2√6/5sin(∏x/a)sin(∏x/a)+2√6/5sin(∏x/a)sin(2∏x/a)+4/5(sin(2∏x/a))^2

but I didn't find the final value. there's still "a" in the final result instead of a number of probability.

I honestly am having a hard time understand what you wrote, please write in Latex, it's there for a reason.

If I understand your question, you're asking how to find the probability that a measurement will produce a value ##E_n##, right?

Assuming that your wave function is already normalized, recall that

$$\Psi (x,0) = \sum ^{\infty} _{n=1} c_n \psi _n (x)$$

What does ##c_n## tell you?
 
  • #7
R-ckay said:
[itex]\int[/itex](boundary a to 0) |ψ(x,0)|^2=[ √6/√5sin(∏x/a)+2/√5sin(2∏x/a)][√6/√5sin(∏x/a)+2/√5sin(2∏x/a]
[itex]\int[/itex] [6/5(sin(∏x/a))^2+2√6/5sin(∏x/a)sin(∏x/a)+2√6/5sin(∏x/a)sin(2∏x/a)+4/5(sin(2∏x/a))^2

but I didn't find the final value. there's still "a" in the final result instead of a number of probability.
The presence of ##a## is not really a problem. What exactly are you trying to do here?
 

FAQ: How Do You Calculate the Probability of Finding a Particle in the Ground State?

What is a "particle in a box" in quantum mechanics?

A particle in a box is a theoretical model used in quantum mechanics to describe a particle confined within a finite space. The particle is assumed to have no interaction with the walls of the box, and its energy is quantized due to the boundary conditions.

How is the probability of finding a particle in a box calculated?

The probability of finding a particle in a particular location within the box is calculated using the Schrödinger equation, which takes into account the particle's wave function and the potential energy of the box. The square of the wave function represents the probability density, which is used to calculate the probability of finding the particle in a specific region.

What is the relationship between the size of the box and the probability of finding the particle?

The probability of finding a particle in a box is inversely proportional to the size of the box. As the box becomes smaller, the energy levels become more closely spaced, leading to a higher probability of finding the particle in a specific region. Conversely, as the box becomes larger, the energy levels become more spread out, resulting in a lower probability of finding the particle in a specific region.

Can the probability of finding a particle in a box ever be 100%?

In theory, the probability of finding a particle in a box can never be 100%. This is because the uncertainty principle in quantum mechanics states that it is impossible to know both the exact position and momentum of a particle simultaneously. However, as the size of the box decreases, the probability of finding the particle within a certain range approaches 100%.

How does the energy of a particle in a box change as the size of the box is altered?

The energy of a particle in a box is directly related to the size of the box. As the size of the box increases, the energy levels also increase. This is due to the fact that a larger box allows for more possible states for the particle to occupy, resulting in a greater range of energy levels. Conversely, a smaller box restricts the possible states and thus leads to lower energy levels.

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