Time evolution of wave function in an infinite square well potential

In summary: So, this is the answer: You are not dividing two numbers that are equal. Instead you are dividing two numbers that are "equivalent". In other words, they are different numbers that have the same magnitude (size) and the same direction in the complex plane. This only happens when you raise both sides of the equation to the same power. In other words, if you have the equation: ##x^n = y^n##, then it's true that ##x = y##. This is a special property of exponentials in the complex plane. It's a consequence of the fact that: ##\exp(z) = \exp(z + 2k\pi i)##
  • #1
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IMG_20181214_131148.jpg

For this problem at t=0
Ψ(x,0)=Ψ13
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?
 

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  • #2
Apashanka said:
View attachment 235705
For this problem at t=0
Ψ(x,0)=Ψ13
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.
 
  • #3
Apashanka said:
(-1)^(-1/8) and (-1)^(-9/8) respectively and the two are equal
These are not quite equal.
 
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  • #4
TSny said:
These are not quite equal.
Will you please explain why??
Thanks
 
  • #5
Apashanka said:
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.
 
  • #6
PeroK said:
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.
 
Last edited:
  • #7
Apashanka said:
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##.
 
  • #8
Apashanka said:
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?
 
  • #9
PeroK said:
##\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??
 
  • #10
TSny said:
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
 

What is the "infinite square well potential"?

The infinite square well potential is a simplified model used in quantum mechanics to describe the behavior of a particle confined to a one-dimensional space. It assumes that the potential energy within the boundaries of the well is constant and infinite, while the potential energy outside the well is zero.

What is the wave function in the infinite square well potential?

The wave function in the infinite square well potential is a mathematical function that describes the probability of finding a particle in a particular location within the well. It is a solution to the Schrodinger equation and takes a sinusoidal form within the boundaries of the well.

What is the time evolution of the wave function in the infinite square well potential?

The time evolution of the wave function in the infinite square well potential describes how the wave function changes over time. It is governed by the Schrodinger equation and shows that the wave function oscillates between its maximum and minimum values as time progresses.

What is the significance of the time evolution of the wave function in the infinite square well potential?

The time evolution of the wave function in the infinite square well potential is significant because it allows us to predict the behavior of a particle in a confined space. It shows that the particle's position becomes more and more certain as time goes on, and it also demonstrates the particle's quantized energy levels.

How does the time evolution of the wave function in the infinite square well potential relate to the Heisenberg uncertainty principle?

The time evolution of the wave function in the infinite square well potential is related to the Heisenberg uncertainty principle because it shows that as the particle's position becomes more certain, its momentum becomes less certain, and vice versa. This is a fundamental principle of quantum mechanics and is represented mathematically by the uncertainty principle equation.

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