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

1. Jul 25, 2014

### Mniazi

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>??

Last edited: Jul 25, 2014
2. Jul 25, 2014

### Jilang

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

3. Jul 25, 2014

### Jilang

A) what is x?

4. Jul 25, 2014

### Mniazi

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.

5. Jul 25, 2014

Staff Emeritus
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>.

6. Jul 25, 2014

### Mniazi

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> ?

7. Jul 26, 2014

### Jilang

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.

8. Jul 26, 2014

### Mniazi

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

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

9. Jul 26, 2014

### Mniazi

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

10. Jul 26, 2014

### Mniazi

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

11. Jul 26, 2014

### Jilang

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......

12. Jul 26, 2014

### Mniazi

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

#### Attached Files:

• ###### Quantum Paper.docx
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13. Jul 26, 2014

### Jilang

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

14. Jul 26, 2014

### Mniazi

sure, here you go, :)

#### Attached Files:

• ###### Quantum Paper.pdf
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15. Jul 26, 2014

### Jilang

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.

16. Jul 26, 2014

### Mniazi

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.

17. Jul 26, 2014

### Jilang

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

18. Jul 26, 2014

### Mniazi

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 .

19. Jul 26, 2014

### Mniazi

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

20. Jul 26, 2014

### Mniazi

FIXED!! here it is updated:

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