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

[R-ckay]

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

 

[vanhees71]

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!

 

[R-ckay]

yes I have, sir

 

[Astrum]

Post your attempt at the solution.

 

[R-ckay]

\int(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]
\int [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.

 

[Astrum]

\int(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]
\int [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?

 

[vela]

\int(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]
\int [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?