Time evolution of wave function in an infinite square well potential

[Apashanka]

For this problem at t=0
Ψ(x,0)=Ψ₁-Ψ₃
Where Ψ₁ and Ψ₃are the normalised eigenstates corresponding to energy level 1 and 3 of the infinite square well potential.
Now for it's time evolution it will be Ψ₁exp(-iE₁t/ħ)- Ψ₃exp(-iE₃t/ħ)
And taking the time given in the question the time part of the individual states is coming to be exp(-iπ/8) and exp(-9iπ/8) respectively which is equivalent to (-1)^(-1/8) and (-1)^(-9/8) respectively and the two are equal
Hence in my opinion the ans is option 3
But the correct option is given option 4??
Am I right?

 

[PeroK]

View attachment 235705
For this problem at t=0
Ψ(x,0)=Ψ₁-Ψ₃
Where Ψ₁ and Ψ₃are the normalised eigenstates corresponding to energy level 1 and 3 of the infinite square well potential.
Now for it's time evolution it will be Ψ₁exp(-iE₁t/ħ)- Ψ₃exp(-iE₃t/ħ)
And taking the time given in the question the time part of the individual states is coming to be exp(-iπ/8) and exp(-9iπ/8) respectively which is equivalent to (-1)^(-1/8) and (-1)^(-9/8) respectively and the two are equal
Hence in my opinion the ans is option 3
But the correct option is given option 4??
Am I right?

You look right to me. Note also that the wavefunctions given are not normalised.

 

[TSny]

(-1)^(-1/8) and (-1)^(-9/8) respectively and the two are equal

These are not quite equal.

 

[Apashanka]

These are not quite equal.

Will you please explain why??
Thanks

 

[PeroK]

Will you please explain why??
Thanks

The problem with $\pi$ is that $\pi$ radians ought to be a full circle. But, it's not. Instead $2\pi$ radians is a full circle. You always have to be careful with this. I make this mistake all the time: especially in the case that $\pi/4$ is not a right angle. I always think $\pi/4$ should be a right angle, but it's not.

There's actually a semi-serious movement in maths education to replace $\pi$ with $\tau$ where $\tau = 2\pi$. Then $\tau$ would be a full circle and $\tau/4$ would be a right angle. I'd move to $\tau$ tomorrow if I could.

Sorry for misleading you. My mistake.

 

[Apashanka]

The problem with $\pi$ is that $\pi$ radians ought to be a full circle. But, it's not. Instead $2\pi$ radians is a full circle. You always have to be careful with this. I make this mistake all the time: especially in the case that $\pi/4$ is not a right angle. I always think $\pi/4$ should be a right angle, but it's not.

There's actually a semi-serious movement in maths education to replace $\pi$ with $\tau$ where $\tau = 2\pi$. Then $\tau$ would be a full circle and $\tau/4$ would be a right angle. I'd move to $\tau$ tomorrow if I could.

Sorry for misleading you. My mistake.

I have used the fact that exp(imπ)={exp(iπ)}^m={cos(π)+i*sin(π)}^m=(-1)^m
Where m can be both positive and negative real number.
Will you please explain whether it is justifiable ,I want to clarify??
Thanks
I have used this fact by which exp(-iπ/8) =(-1)^(-1/8) and exp (-9iπ/8)=(-1)^(-9/8) and the two are equal.

 

[PeroK]

I have used the fact that exp(imπ)={exp(iπ)}^m={cos(π)+i*sin(π)}^m=(-1)^m
Where m can be both positive and negative real number.
Will you please explain whether it is justifiable ,I want to clarify??
Thanks
I have used this fact by which exp(-iπ/8) =(-1)^(-1/8) and exp (-9iπ/8)=(-1)^(-9/8) and the two are equal.

$\exp(\frac{-i\pi}{8}) = \exp(\frac{-i17\pi}{8}) \ne \exp(\frac{-i9\pi}{8})$

The period of $\exp, \sin, \cos$ is $2 \pi$, not $\pi$.

 

[TSny]

I have used this fact by which exp(-iπ/8) =(-1)^(-1/8) and exp (-9iπ/8)=(-1)^(-9/8) and the two are equal.

What do you get if you divide exp (-iπ/8) by exp(-9iπ/8)? First reduce exp (-iπ/8) $\div$ exp(-9iπ/8) to a single exponential. Likewise, what does (-1)^(-1/8) $\div$ (-1)^(-9/8) reduce to?

 

[Apashanka]

$\exp(\frac{-i\pi}{8}) = \exp(\frac{-i17\pi}{8}) \ne \exp(\frac{-i9\pi}{8})$

The period of $\exp, \sin, \cos$ is $2 \pi$, not $\pi$.

Ohh thanks sir for the clarification .
But sir will you please explain how I am getting the two to be same which I have posted above that remains a confusion??

 

[Apashanka]

What do you get if you divide exp (-iπ/8) by exp(-9iπ/8)? First reduce exp (-iπ/8) $\div$ exp(-9iπ/8) to a single exponential. Likewise, what does (-1)^(-1/8) $\div$ (-1)^(-9/8) reduce to?

Yes sir that's coming to be -1
Thanks sir