Question on Quantum Physics- Probability of finding a particle

[warhammer]

Homework Statement

Consider two states of a particle given by the wave functions
ψ1(x) = √re-|r|x and ψ2(x) = -√re-|r|x
The probability of finding the particle in the range -1/r < x < 1/r is p1 in the first state and p2 in the second state. Which of the following is appropriate?
(a) p1 = p2
(b) p1 = - p2
(c) p1 < p2
(d) p1 > p2

Relevant Equations

Integration {ψ(x)}{ψ(x)*} from -∞ to ∞=1
(* denotes conjugate)

I calculated the complex conjugate of both the given wavefunctions. For ψ1: ∫re^((-2)mod(r)x)dx=1 with upper limit ∞ & lower limit -∞. I replaced the upper and lower limit after breaking down the function inside integration as follows- r*∫e^(2rx)dx from -1/r to 0 and r*e∫e^(-2rx)dx from 0 to 1/r. The answer was 1-1/e^2 which equals 0.86.

I repeated the above steps for ψ2 and similarly obtained 0.86. However, I am somewhat not sure if I have proceeded correctly.

 

[PeroK]

Homework Statement: Consider two states of a particle given by the wave functions
ψ1(x) = √re-|r|x and ψ2(x) = -√re-|r|x
The probability of finding the particle in the range -1/r < x < 1/r is p1 in the first state and p2 in the second state. Which of the following is appropriate?
(a) p1 = p2
(b) p1 = - p2
(c) p1 < p2
(d) p1 > p2
Homework Equations: Integration {ψ(x)}{ψ(x)*} from -∞ to ∞=1
(* denotes conjugate)

I calculated the complex conjugate of both the given wavefunctions. For ψ1: ∫re^((-2)mod(r)x)dx=1 with upper limit ∞ & lower limit -∞. I replaced the upper and lower limit after breaking down the function inside integration as follows- r*∫e^(2rx)dx from -1/r to 0 and r*e∫e^(-2rx)dx from 0 to 1/r. The answer was 1-1/e^2 which equals 0.86.

I repeated the above steps for ψ2 and similarly obtained 0.86. However, I am somewhat not sure if I have proceeded correctly.

Those wavefunctions look kinda simlilar to me.

 

[warhammer]

Those wavefunctions look kinda simlilar to me.

Yes. That is what I figured because when we find the complex conjugate, the negative sign for the second one would vanish and the functions on closer inspection would give us the same values upon integrating on the said limits. I found out the answer to be p1=p2 but I'm somewhat new to these concepts and confused whether my approach is correct or not. I hope this doesn't break the rules of the forum but since I have done the solution part myself would you be able to atleast guide me if I'm correct here or not?

 

[PeroK]

Yes. That is what I figured because when we find the complex conjugate, the negative sign for the second one would vanish and the functions on closer inspection would give us the same values upon integrating on the said limits. I found out the answer to be p1=p2 but I'm somewhat new to these concepts and confused whether my approach is correct or not. I hope this doesn't break the rules of the forum but since I have done the solution part myself would you be able to atleast guide me if I'm correct here or not?

The two wavefunctions are essentially equivalent in terms of measurements, probabilities and expectation values. You can see that as soon as you set up the two integrals, they are the same.

Any two wavefunctions that differ only by a "phase factor" - any complex number of unit modulus (in this case $-1$) - have this property of returning the same probabilities. Again, you can see that as soon as you set up the integral.

 

[warhammer]

The two wavefunctions are essentially equivalent in terms of measurements, probabilities and expectation values. You can see that as soon as you set up the two integrals, they are the same.

Any two wavefunctions that differ only by a "phase factor" - any complex number of unit modulus (in this case $-1$) - have this property of returning the same probabilities. Again, you can see that as soon as you set up the integral.

Thank you for your help, got this one now :)