Average of Momentum for 1D Quantum Harmonic Oscillator

In summary, when solving for the expectation and variance of a 1D QHO using Dirac operators, it is necessary to take into account the index of state and the parity of the state. If the index of state is changed, the result will be wrong.
  • #1
LizardWizard
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For a 1D QHO we are given have function for ##t=0## and we are asked for expectation and variance of P at some time t.

##|\psi>=(1/\sqrt 2)(|n>+|n+1>)## Where n is an integer

So my idea was to use Dirac operators ##\hat a## and ##\hat a^\dagger## and so I get the following solution

##<\hat P>=<\psi|\hat P|\psi>=i \sqrt {m\omega\hbar/2} <\psi|(\hat a^\dagger - \hat a^)|\psi>=##
##=i \sqrt {m\omega\hbar/2}(1/\sqrt 2) (1/\sqrt 2)(<n|+<n+1|(\hat a^\dagger-\hat a)|n>+|n+1>)##

Knowing that
##\hat a^\dagger |\psi_n>= \sqrt {n+1} |\psi_{n+1}>##
##\hat a |\psi_n>= \sqrt {n} |\psi_{n-1}>##


I eventually arrived at

##<\hat P>=i\sqrt{m\omega\hbar/8}(-\sqrt n <n|n> + \sqrt{n+1} <n+1|n+1>)##

However the answer is supposed to be 0, did I do something wrong or could I perhaps make use of the fact that ##\sqrt{n+1}-\sqrt n ## for ##n\to\infty## equal to 0?
 
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  • #2
It seems that you made a mistake when calculating ##\hat{a}|n+1\rangle##, what value did you get for this?
 
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  • #3
blue_leaf77 said:
It seems that you made a mistake when calculating ##\hat{a}|n+1\rangle##, what value did you get for this?
That would be ##\sqrt n |n>## but I think I did forget the terms for ##\hat a^\dagger|n+1>## and <n+1| which would be ##\sqrt{n+2}|n+2>## and the corresponding bra.
 
  • #4
LizardWizard said:
That would be ##\sqrt n |n>##.
No, that's wrong. So, it's clear now that this is at least one of the mistakes.
LizardWizard said:
but I think I did forget the terms for ##\hat a^\dagger|n+1>##
The above expression will give rise to a term proportional to ##|n+2\rangle## in the right part of the braket. Meanwhile in the left part of the braket there is no ##\langle n+2|##. This means, you can safely abandon the term ##\hat a^\dagger|n+1>##.
 
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  • #5
I knew there was a reason I didn't include the |n+2> term in the first place. However what exactly is wrong about that other result?
It follows from applying ##\hat a |\psi_n>= \sqrt {n} |\psi_{n-1}>## to |n+1>
 
  • #6
LizardWizard said:
I knew there was a reason I didn't include the |n+2> term in the first place. However what exactly is wrong about that other result?
It follows from applying ##\hat a |\psi_n>= \sqrt {n} |\psi_{n-1}>## to |n+1>
No, it doesn't. For instance, note that the index of the state in the left hand side is the same as the thing under the square root in the right hand side. Now what happen to this thing if the index of state in the left hand side is changed to ##n+1##?
 
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  • #7
Oh I see, I found it weird because I have an example from class that makes uses it the way I said, but in the case we only had two possible n-states 0 and 1 instead of a generic integer n, I suposse that's what makes the difference.

Okay, I redid the calculations and this pretty much breaks all parity (except for one that canceled itself) and the result is 0. Thank you for the help.
 

What is the concept of momentum in a 1D quantum harmonic oscillator?

In a 1D quantum harmonic oscillator, momentum refers to the product of an object's mass and velocity. In the context of quantum mechanics, momentum is described by the operator p, which acts on the wave function of a particle to determine its momentum state.

How is the average of momentum calculated for a 1D quantum harmonic oscillator?

The average of momentum for a 1D quantum harmonic oscillator is calculated by taking the expectation value of the momentum operator, p, over the wave function of the system. This is done by integrating the wave function multiplied by the conjugate of the momentum operator over all possible positions.

How does the average of momentum change as the energy of the system increases?

As the energy of a 1D quantum harmonic oscillator increases, the average of momentum also increases. This is because higher energy levels correspond to larger spatial fluctuations in the wave function, resulting in a wider spread of momentum values.

What is the significance of the uncertainty principle in relation to the average momentum of a 1D quantum harmonic oscillator?

According to the uncertainty principle, it is impossible to know the exact position and momentum of a particle simultaneously. This means that the average of momentum for a 1D quantum harmonic oscillator will always have some degree of uncertainty associated with it. As the uncertainty in position decreases, the uncertainty in momentum increases and vice versa.

How does the average momentum for a 1D quantum harmonic oscillator compare to that of a classical harmonic oscillator?

The average momentum for a 1D quantum harmonic oscillator is quantized, meaning it can only take on discrete values, while the average momentum for a classical harmonic oscillator can take on any value. Additionally, the average momentum for a quantum harmonic oscillator is dependent on the energy level of the system, while for a classical harmonic oscillator, the average momentum is directly proportional to the displacement of the object from its equilibrium position.

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