# What is <x_1-x_2> for two particles in a 1-D harmonic oscillator

## Main Question or Discussion Point

if we have two non-interacting particles of mass M in a one-dimensional harmonic oscillator potential of frequency ω, with the wavefunction defined as:

$$\Psi\left(x_1,x_2\right) = \psi_n\left(x_1\right) \psi_m\left(x_2\right)$$

where x_1 and x_2 are two particle co-ordinates. and ψ_n is the nth harmonic oscillator eigenfunction.

then:
a) will:
$$\psi_n(x_1)= (\frac{\frac{M*\omega}{\hbar}}{\pi})^{1/4}*H_n(x_1)*e^{-\frac{M*x_1^2*\omega}{2*h}}$$

or will it be in this format:

$$\frac{1}{\sqrt{2^n*n!}}*(\frac{m*\omega}{\pi*\hbar})^{1/4}*e^{-\frac{M*x_1^2*\omega}{2*\hbar}}*H_n(\sqrt{\frac{m*\omega}{\hbar}}*x)$$

b) What is <x_1-x_2>??

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## Answers and Replies

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B) Would it be the same as <x_1> - <x_2> as they are non-interacting?

A) what is x?

Since im new to quantum physics I want to even know what <x_1> represents. sorry that had to be x_1 or x_2 in the second formulas, these are the positions of the two particles.

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I think you need to think harder about what it is you are trying to calculate. <x1 - x2> should be zero. <|x1 - x2|> won't be zero, nor will sqrt(<(x1-x2)^2>.

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Can you please tell me what something inside < > represents? Like the thing im trying to solve needs me to calculate <(x_1-x_2)^2> ?

It means the expectation value. (What it might be found to be on average.) So <x1-x2> is the expectation value of the vector distance between the two particles and <(x1-x2)^2> is the expectation value of the square of the distance between the two particles. The first one will be zero as the first particle will expected to be the left of the second particle as often as it is the right of it etc.

1 person
Thanks, So incase I wanted to make sure that <x_1-x_2> is zero then will I have to do this integral? :

$$<(x_1-x_2)> = \int_{-\inf}^{\inf}{(\psi_n^{*}(x_1)*x_1*\psi_n(x_1))dx_1} - \int_{-\inf}^{\inf}{(\psi_m^{*}(x_2)*x_2*\psi(x_2))dx_2}$$

And If I want to calculate $$<(x_1-x_2)^2>$$ then can i write it as $$<x_1^2>-<2*x_1><x_2>+<x_2^2>$$ ?

Well they are non-interacting particles, and when I solve these I do get 0.

Thanks, So incase I wanted to make sure that <x_1-x_2> is zero then will I have to do this integral? :

$$<(x_1-x_2)> = \int_{-\inf}^{\inf}{(\psi_n^{*}(x_1)*x_1*\psi_n(x_1))dx_1} - \int_{-\inf}^{\inf}{(\psi_m^{*}(x_2)*x_2*\psi(x_2))dx_2}$$
Yes, I'm pretty sure it will work that way if x1 and x2 are independent. I haven't studied this though, so I am surmising......

Here is the work I did, two pages full of calculations, am I correct?

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Mniazi, can you send PDF, I'm on an iPad here?

sure, here you go, :)

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Thanks! The last equation on the first page is not right. You need to sandwich x^2 between the ψ s in the integral, rather than x and squaring the whole thing.

so the <x_1^2> and <x_2^2> integrals are wrong. Oh! ok, do you mean to say that the wavefunctions should stay as they are, and only x_1 should be changed to x_1^2 respectively? Thanks! I shall correct this imediately.

Yes, and then you don't need to square the whole thing.

When I do the integral from -infinity to infinity, it doesnt converge? Edit: I think there is a problem with my formula, wait, let me fix it .

yeah, in the doc I forgot to put a square over the x in the wavefunction exponent.

FIXED!! here it is updated:

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here is the more corrected one, I mistakenly redirected M as the wavefunction modulus with variable, which could cause confusion with the Mass

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Good morning Mniazi. What is H(y)? Aren't these the hermite polynomials? I think they should be a function of x.

Yes! These are Hermite Polynomials, but since theey are a function of $$\sqrt{\frac{M*\omega}{\pi}*x}$$, so I wrote it with respect to y, which the document states equals to $$\sqrt{\frac{M*\omega}{\pi}*x}$$. :)

OK, I missed that. If that's the case why are they appearing in your final expressions? Shouldn't they be evaluated as part of the integral?

I even checked the integration on Mathematica, It is coming the same. should they not appear in the final equation?