Proving Delta x^2 & Delta p^2 for Harmonic Oscillator

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Homework Help Overview

The discussion revolves around proving that neither \(\Delta x\) nor \(\Delta p\) is generally constant for a one-dimensional harmonic oscillator. Participants are tasked with demonstrating that \((\Delta x)^2\) and \((\Delta p)^2\) take a specific form involving cosine functions, where \(\omega\) is the oscillator's frequency.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants express uncertainty about the meaning of \(\Delta\) and its implications for the problem's setup. Some suggest that \(\Delta x\) should follow a specific functional form, while others discuss the need to calculate expectation values using wavefunctions expressed as superpositions of states. Questions arise regarding the evaluation of integrals and the nature of the wavefunctions involved.

Discussion Status

The conversation is ongoing, with various participants exploring different interpretations of the problem. Some have proposed that the integrals should be treated as parameters rather than computed directly. Others have introduced relevant concepts from quantum mechanics, such as matrix elements and the use of raising and lowering operators, while acknowledging gaps in their current coursework.

Contextual Notes

Participants note that the problem may not align with their current curriculum, leading to confusion about certain concepts and methods. There is a recognition that some foundational topics, like the properties of Hermite polynomials, have not been covered in class.

stunner5000pt
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Homework Statement


Show taht neither [itex]\Delta x[/itex] nor [itex]\Delta p[/itex] is generally constant (independent of time) for a general state of the one dimensional harmonic oscillator. Prove that [itex](\Delta x)^2[/itex] and [itex](\Delta p)^2[/itex] are both of the form

[tex](\Delta)^2 = A + B \cos^2 \omega t[/tex]
where omega is the frequency associated with the oscillator.

2. The attempt at a solution

First of all I am not really sure wht the Delta means. Does it mean Delta x should have that form??

something like

[tex](\Delta x)^2 = A + B \cos^2 \omega t[/tex]
where A and B are some constants??

this is where i think is a logical beginning to th solution

since we are talking about the genferla case of the harmonic oscillator then the wavefunction must be written as a superposition of states??

[tex]\Psi(x,t) = \sum_{n=0}^{\infty} c_{n} \psi_{n}(x) e^{-iE_{n}t/\hbar}[/tex]

so we can calculate the expectation value of x
[tex]\left<x(t)\right> =\int_{\infty}^{\infty} \Psi(x,t)^* x \Psi(x,t) dx[/tex]

and [tex]\left<(x(t))^2\right> =\int_{\infty}^{\infty} \Psi(x,t)^* x^2 \Psi(x,t) dx[/tex]

we're going to get cross terms like

[tex]\left<x\right> = \int \psi_{m}^* x \psi_{n} dx[/tex]
[tex]\left<x^2\right> =\int c_{m}^* c_{n} x_{m,n} \exp\left(\frac{iE_{m,n} t}{\hbar}\right)[/tex]

now I am just wondering how to evalue these integrals

thanks for any help!
 
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stunner5000pt said:
First of all I am not really sure wht the Delta means. Does it mean Delta x should have that form??

[tex]\Delta x = \sqrt{|<x>^2-<x^2>|}[/tex]

I put the absolute value bars in there because I forget the actual order of subtraction. :redface:

now I am just wondering how to evalue these integrals

Do you know what the wavefunctions look like? The spatial part of the Harmonic oscillator wavefunctions are Hermite polynomials, modulated by a Gaussian exponential function. Does that help?
 
Tom Mattson said:
[tex]\Delta x = \sqrt{|<x>^2-<x^2>|}[/tex]

I put the absolute value bars in there because I forget the actual order of subtraction. :redface:



Do you know what the wavefunctions look like? The spatial part of the Harmonic oscillator wavefunctions are Hermite polynomials, modulated by a Gaussian exponential function. Does that help?

hmmm we never actually went into determining these in class so I am just wondering

its not in the textbook either

but i can acquire the formula for the hermite polynomials and get an answer from there i think
 
stunner5000pt said:

Homework Statement


Show taht neither [itex]\Delta x[/itex] nor [itex]\Delta p[/itex] is generally constant (independent of time) for a general state of the one dimensional harmonic oscillator. Prove that [itex](\Delta x)^2[/itex] and [itex](\Delta p)^2[/itex] are both of the form

[tex](\Delta)^2 = A + B \cos^2 \omega t[/tex]
where omega is the frequency associated with the oscillator.

2. The attempt at a solution

First of all I am not really sure wht the Delta means. Does it mean Delta x should have that form??

something like

[tex](\Delta x)^2 = A + B \cos^2 \omega t[/tex]
where A and B are some constants??

this is where i think is a logical beginning to th solution

since we are talking about the genferla case of the harmonic oscillator then the wavefunction must be written as a superposition of states??

[tex]\Psi(x,t) = \sum_{n=0}^{\infty} c_{n} \psi_{n}(x) e^{-iE_{n}t/\hbar}[/tex]

so we can calculate the expectation value of x
[tex]\left<x(t)\right> =\int_{\infty}^{\infty} \Psi(x,t)^* x \Psi(x,t) dx[/tex]

and [tex]\left<(x(t))^2\right> =\int_{\infty}^{\infty} \Psi(x,t)^* x^2 \Psi(x,t) dx[/tex]

we're going to get cross terms like

[tex]\left<x\right> = \int \psi_{m}^* x \psi_{n} dx[/tex]
[tex]\left<x^2\right> =\int c_{m}^* c_{n} x_{m,n} \exp\left(\frac{iE_{m,n} t}{\hbar}\right)[/tex]

now I am just wondering how to evalue these integrals

thanks for any help!


The question does not quite make sense to me. If you take a really general state, you won't get an expression lik ethey give but rather a sum of several cosine terms all with different values of angular frequency so I am confused by that question.

This siad, what you should do is NOT carry out the integrations but give them a name "A" or "B" and so on. All you need to know is that those are real parameters. You will find that some of those constants will multiply complex exponentials but the complex exponentials will come in pairs which wil be complex conjugate of each other. Using [itex]e^{i \theta} + e ^{-i \theta} = 2 cos (\theta)[/itex] you will end up with a bunch of cosine of [itex]\omega t[/itex]. But again, to get only one cos in your final answer, you woul dneed to consider a superposition of only two states of different energies, not more than that.
 
There are analytic expressions for the matrix elements <m|x|n> and
<m|x^2|n> for the H.O.

In fact the elements <m|x|n> come up in the Fermi's golden rule treatment of infrared radiation causing transitions from state |m> to state |n>.

For anyone who's interested

[itex]<i|\hat{x}|j>=\delta_{j,i-1}\sqrt{(j+1)\hbar\<br /> /2m\omega_0}+\delta_{j,i+1}\sqrt{j\hbar/2m\omega_0}$[/itex]
 
christianjb said:
There are analytic expressions for the matrix elements <m|x|n> and
<m|x^2|n> for the H.O.

In fact the elements <m|x|n> come up in the Fermi's golden rule treatment of infrared radiation causing transitions from state |m> to state |n>.

For anyone who's interested

[itex]<i|\hat{x}|j>=\delta_{j,i-1}\sqrt{(j+1)\hbar\<br /> /2m\omega_0}+\delta_{j,i+1}\sqrt{j\hbar/2m\omega_0}$[/itex]

Of course!:rolleyes: For some reason I thought this was about the infinite square well.

So indeed, using raising and lowering operators, One may find the expectation values of x and x^2 in an arbitrary state. The x will connect two states differing by one in the quantum number "n". The x^2 will connect two states of same n or differing by two. One can write the final result in terms of the harmonic oscillator omega.
Sorry for my mistake :frown:
 
nrqed said:
Of course!:rolleyes: For some reason I thought this was about the infinite square well.

So indeed, using raising and lowering operators, One may find the expectation values of x and x^2 in an arbitrary state. The x will connect two states differing by one in the quantum number "n". The x^2 will connect two states of same n or differing by two. One can write the final result in terms of the harmonic oscillator omega.
Sorry for my mistake :frown:

There were two posts yesterday on QM wells. The other was asking for help on the infinite square well- hence the confusion. For some reason- I mixed them up in my mind too at first.
 
nrqed said:
Of course!:rolleyes: For some reason I thought this was about the infinite square well.

So indeed, using raising and lowering operators, One may find the expectation values of x and x^2 in an arbitrary state. The x will connect two states differing by one in the quantum number "n". The x^2 will connect two states of same n or differing by two. One can write the final result in terms of the harmonic oscillator omega.
Sorry for my mistake :frown:

hmm i have seen the solution of the expectation value of x and x^2 using the ladder operators but we haven't covered that in class... and we won't in this course

but i will learn about this during the summer :smile:
 

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