Spatial Wave function of two indistinguishable particles

[Zero1010]

Homework Statement

Spatial wave function for two particles

Relevant Equations

$\psi^{+}(x_{1}, x_{2})=\frac{1}{\sqrt2}[\psi_{n}(x_{1})\psi_{k}(x_{2})+\psi_{k}(x_{1})\psi_{n}(x_{2}))$

$V_{A}(x) = \sqrt\frac{2}{L} sin (\frac{\pi x}{L})$

$V_{B}(x) = \sqrt\frac{2}{L} sin (\frac{2\pi x}{L})$

Hi,

I just need someone to check over my work. I am having trouble with the next part of this question and I just wanted to check that this part was correct first.

I have two particles in an infinite square well (walls at x=0 and x=L). I need write an expression for the spatial wave function. From the question I know that the spatial part is symmetric (spin is antisymmetric and the particles are fermions) so will be given by

$\psi^{+}(x_{1}, x_{2})=\frac{1}{\sqrt2}[\psi_{n}(x_{1})\psi_{k}(x_{2})+\psi_{k}(x_{1})\psi_{n}(x_{2}))$

One of the particles is in the ground state and the other is in the first excited stat. Normalised eigenfunctions given in the question are:

$V_{A}(x) = \sqrt\frac{2}{L} sin (\frac{\pi x}{L})$

$V_{B}(x) = \sqrt\frac{2}{L} sin (\frac{2\pi x}{L})$

Therefore for the spatial part I have:

$\psi^{+}(x_{1}, x_{2})=\frac{1}{\sqrt2} [\sqrt\frac{2}{L} sin (\frac{\pi x_{1}}{L})\sqrt\frac{2}{L} sin (\frac{2\pi x_{2}}{L}) + \sqrt\frac{2}{L} sin (\frac{2\pi x_{1}}{L}) \sqrt\frac{2}{L} sin (\frac{\pi x_{2}}{L})]$

Which gives the spatial wavefunction:
$\psi^{+}(x_{1}, x_{2})=\frac{1}{\sqrt L} [ sin (\frac{\pi x_{1}}{L}) sin (\frac{2\pi x_{2}}{L}) + sin (\frac{2\pi x_{1}}{L})sin (\frac{\pi x_{2}}{L})]$

Does this make sense or am I missing something? Can I simplify this further?

Thanks in advance for help. This forum has been amazing.

Thanks

 

[PeroK]

Looks right to me.

 

[Zero1010]

Cool thanks. I'll have another look at the second part of the question tomorrow.

 

[Zero1010]

Ok if possible it would be great to check over my work here and maybe some advice on how to proceed as I've hit a brick wall with some integration.

In the next part of the question I have two indistignishable particles in an infinite square well with walls at x=0 and x=L.

Firstly, I need to write an expression from the probability density that one particle will be found at $x_{1}$ and the other particle found at $x_{2}$.

Heres my work so far -

To get the probability density I take the square of the modulus:

probability density=$\left| \psi(x_{1}, x_{2}) \right|^{2}$

probability density=$\left| \frac{L}{\sqrt 2}[sin( \frac{\pi x_{1}}{L})sin( \frac{2\pi x_{2}}{L})+sin( \frac{2\pi x_{1}}{L})sin( \frac{\pi x_{2}}{L}) \right|^{2}$

Which gives:

probability density=$\frac{L^{2}}{2}[sin^{2}( \frac{\pi x_{1}}{L})sin^{2}( \frac{2\pi x_{2}}{L})+sin( \frac{\pi x_{1}}{L})sin( \frac{2\pi x_{2}}{L})sin( \frac{2\pi x_{1}}{L})sin( \frac{\pi x_{2}}{L})+sin^{2}( \frac{2\pi x_{1}}{L})sin^{2}( \frac{\pi x_{2}}{L})]$

Hopefully that looks good up to now but I am a bit worried I have made an error squaring this.

The next part of the question asks to calculate the probability that both particles are found in the left hand side of the well at $0\leq x_{1} \lt \frac{L}{2}$ and $0\leq x_{2} \lt \frac{L}{2}$

To get the probability I need to integrate the probability density over $x_{1}$ and then over $x_{2}$:

$\frac{L^{2}}{2}\int_0^\frac{L}{2} \int_0^\frac{L}{2}[sin^{2}( \frac{\pi x_{1}}{L})sin^{2}( \frac{2\pi x_{2}}{L})+sin( \frac{\pi x_{1}}{L})sin( \frac{2\pi x_{2}}{L})sin( \frac{2\pi x_{1}}{L})sin( \frac{\pi x_{2}}{L})+sin^{2}( \frac{2\pi x_{1}}{L})sin^{2}( \frac{\pi x_{2}}{L})]dx_{1}, dx_{2}$

This is where I have hit a brick wall. I think there is a way to separate this into two integrals but I am not sure how to approach this. I'm also not confident that I have done this in the correct order and should have done something with the integrals before squaring for the probability density.

I have been given three integrals in the question to make it easier but I am struggling with how to use them since I don't have any terms to match and I don't think I can just sub them into the equation above:

$\int_0^\frac{L}{2} sin^{2}(\frac{\pi x}{L}) dx = \frac{L}{4}$
$\int_0^\frac{L}{2} sin^{2}(\frac{2\pi x}{L}) dx = \frac{L}{4}$
$\int_0^\frac{L}{2} sin(\frac{\pi x}{L}) sin(\frac{2\pi x}{L} dx = \frac{2L}{3\pi}$

Any help or advice with this would be greatly appreciated.

Thanks

 

[PeroK]

Ok if possible it would be great to check over my work here and maybe some advice on how to proceed as I've hit a brick wall with some integration.

In the next part of the question I have two indistignishable particles in an infinite square well with walls at x=0 and x=L.

Firstly, I need to write an expression from the probability density that one particle will be found at $x_{1}$ and the other particle found at $x_{2}$.

Heres my work so far -

To get the probability density I take the square of the modulus:

probability density=$\left| \psi(x_{1}, x_{2}) \right|^{2}$

probability density=$\left| \frac{L}{\sqrt 2}[sin( \frac{\pi x_{1}}{L})sin( \frac{2\pi x_{2}}{L})+sin( \frac{2\pi x_{1}}{L})sin( \frac{\pi x_{2}}{L}) \right|^{2}$

Which gives:

probability density=$\frac{L^{2}}{2}[sin^{2}( \frac{\pi x_{1}}{L})sin^{2}( \frac{2\pi x_{2}}{L})+sin( \frac{\pi x_{1}}{L})sin( \frac{2\pi x_{2}}{L})sin( \frac{2\pi x_{1}}{L})sin( \frac{\pi x_{2}}{L})+sin^{2}( \frac{2\pi x_{1}}{L})sin^{2}( \frac{\pi x_{2}}{L})]$

Hopefully that looks good up to now but I am a bit worried I have made an error squaring this.

First, you've got a factor of $\frac{L^2}{2}$ there. That doesn't look right. Neither does your quadratic expansion: $(a+ b)^2 = a^2 + 2ab + b^2$.

The next part of the question asks to calculate the probability that both particles are found in the left hand side of the well at $0\leq x_{1} \lt \frac{L}{2}$ and $0\leq x_{2} \lt \frac{L}{2}$

To get the probability I need to integrate the probability density over $x_{1}$ and then over $x_{2}$:

$\frac{L^{2}}{2}\int_0^\frac{L}{2} \int_0^\frac{L}{2}[sin^{2}( \frac{\pi x_{1}}{L})sin^{2}( \frac{2\pi x_{2}}{L})+sin( \frac{\pi x_{1}}{L})sin( \frac{2\pi x_{2}}{L})sin( \frac{2\pi x_{1}}{L})sin( \frac{\pi x_{2}}{L})+sin^{2}( \frac{2\pi x_{1}}{L})sin^{2}( \frac{\pi x_{2}}{L})]dx_{1}, dx_{2}$

This is where I have hit a brick wall. I think there is a way to separate this into two integrals but I am not sure how to approach this. I'm also not confident that I have done this in the correct order and should have done something with the integrals before squaring for the probability density.

Note that in general:
$$\int \int f(x)g(y) dx dy = \bigg ( \int f(x)dx \bigg ) \bigg ( \int g(y) dy \bigg ) $$

 

[Zero1010]

You're a life saver. Its amazing how it sometimes only takes one or two points to open a question up. I think I have it now.

Thanks again

 

[PeroK]

You're a life saver. Its amazing how it sometimes only takes one or two points to open a question up. I think I have it now.

Thanks again

Post your answer once you have it. See whether it agrees with mine!

 

[Zero1010]

Ok, fingers crossed:

For the quadratic expansion I now get (fixing the factor and 2 in front of the second term):

$\frac{1}{L}[sin^{2}(\frac{\pi x_{1}}{L})sin^{2}(\frac{2\pi x_{2}}{L})+2sin(\frac{\pi x_{1}}{L})sin(\frac{\pi x_{2}}{L})sin(\frac{2\pi x_{1}}{L})sin(\frac{2\pi x_{2}}{L})+sin^{2}(\frac{\pi x_{2}}{L})sin^{2}(\frac{2\pi x_{1}}{L})$

The integral becomes:
$\frac{1}{L}\int_0^\frac{L}{2}\int_0^\frac{L}{2} [sin^{2}(\frac{\pi x_{1}}{L})sin^{2}(\frac{2\pi x_{2}}{L})+2sin(\frac{\pi x_{1}}{L})sin(\frac{\pi x_{2}}{L})sin(\frac{2\pi x_{1}}{L})sin(\frac{2\pi x_{2}}{L})+sin^{2}(\frac{\pi x_{2}}{L})sin^{2}(\frac{2\pi x_{1}}{L}) dx_{1}, dx_{2}$

$\frac{1}{L} \int_0^\frac{L}{2}sin^{2}(\frac{\pi x_{1}}{L})\int_0^\frac{L}{2}sin^{2}(\frac{2\pi x_{2}}{L})+2\int_0^\frac{L}{2}sin(\frac{\pi x_{1}}{L})sin(\frac{2\pi x_{1}}{L})\int_0^\frac{L}{2}sin(\frac{\pi x_{2}}{L})sin(\frac{2\pi x_{2}}{L})+\int_0^\frac{L}{2}sin^{2}(\frac{\pi x_{2}}{L})\int_0^\frac{L}{2}sin^{2}(\frac{2\pi x_{1}}{L})$

Now using the integrals given in the question I have
$(\frac{L}{4})(\frac{L}{4})+2(\frac{2L}{3\pi})(\frac{2L}{3\pi})+(\frac{L}{4})(\frac{L}{4})$

$\frac{L^{2}}{8}\frac{4L^{2}}{9\pi^{2}}$

 

[PeroK]

What is your answer?

 

[Zero1010]

Sorry, missed the last bit:

$\frac{L^{4}}{18\pi^{2}}$

 

[PeroK]

Sorry, missed the last bit:

$\frac{L^{4}}{18\pi^{2}}$

That's gone awry at the end. It can't have $L$ in it. For one thing a probability is dimensionless. And I think you've multiplied instead of added some terms. It looked correct until the last step or two.

 

[Zero1010]

Ok thanks. I'll take another look at it.

 

[PeroK]

Here's a tip. Although maybe it's something of a question of style. In QM you will often get sets of eigenfunctions of various descriptions. Sometimes you need to express them fully to get the integrals. But, it's often simpler to keep them as $\psi_1, \psi_2$ etc. as much as you can.

In this case, I would do something like:
$$|\psi(x_1, x_2)|^2 = \frac 1 2 [\psi_1(x_1)^2 \psi_2(x_2)^2 + \psi_2(x_1)^2 \psi_1(x_2)^2 + 2\psi_1(x_1)\psi_2(x_1)\psi_1(x_2)\psi_2(x_2)]$$
$$\int_0^{\frac L 2}\psi_1(x)^2 dx = \frac 2 L \int_0^{\frac L 2}\sin^2(\frac{\pi x}{L}) dx = \frac 2 L (\frac L 4) = \frac 1 2$$
Etc.

 

[Zero1010]

Thanks for the tip, makes it all a lot clearer and easier to set out.

Thanks for all your help so far. I'm going to have to leave it for now and come back to it tomorrow but I'm happy I'm much closer to a solution now.

 

[Zero1010]

I've had a bit of time this morning to look at this and I have had another go. Here is my attempt:

$\frac{1}{L}\int_0^\frac{L}{2} sin^{2}(\frac{\pi x_{1}}{L})dx_{1} = \frac{1}{L} (\frac{L}{4}) = \frac{1}{4}$
$\frac{1}{L}\int_0^\frac{L}{2} sin^{2}(\frac{2\pi x_{2}}{L})dx_{2} = \frac{1}{L} (\frac{L}{4}) = \frac{1}{4}$
$\frac{2}{L}\int_0^\frac{L}{2} sin(\frac{\pi x_{1}}{L}) sin(\frac{2\pi x_{1}}{L})dx_{1} = \frac{2}{L} (\frac{2L}{3\pi}) = \frac{4}{3\pi}$
$\frac{2}{L}\int_0^\frac{L}{2} sin(\frac{\pi x_{2}}{L}) sin(\frac{2\pi x_{2}}{L})dx_{2} = \frac{2}{L} (\frac{2L}{3\pi}) = \frac{4}{3\pi}$
$\frac{1}{L}\int_0^\frac{L}{2} sin^{2}(\frac{\pi x_{2}}{L})dx_{2} = \frac{1}{L} (\frac{L}{4}) = \frac{1}{4}$
$\frac{1}{L}\int_0^\frac{L}{2} sin^{2}(\frac{2\pi x_{1}}{L})dx_{1} = \frac{1}{L} (\frac{L}{4}) = \frac{1}{4}$

Putting this all together gives:

$\frac{2}{16}+\frac{16}{9\pi^{2}} = 0.31$

Hopefully I haven't made any silly mistakes...

 

[PeroK]

I've had a bit of time this morning to look at this and I have had another go. Here is my attempt:

$\frac{1}{L}\int_0^\frac{L}{2} sin^{2}(\frac{\pi x_{1}}{L})dx_{1} = \frac{1}{L} (\frac{L}{4}) = \frac{1}{4}$
$\frac{1}{L}\int_0^\frac{L}{2} sin^{2}(\frac{2\pi x_{2}}{L})dx_{2} = \frac{1}{L} (\frac{L}{4}) = \frac{1}{4}$
$\frac{2}{L}\int_0^\frac{L}{2} sin(\frac{\pi x_{1}}{L}) sin(\frac{2\pi x_{1}}{L})dx_{1} = \frac{2}{L} (\frac{2L}{3\pi}) = \frac{4}{3\pi}$
$\frac{2}{L}\int_0^\frac{L}{2} sin(\frac{\pi x_{2}}{L}) sin(\frac{2\pi x_{2}}{L})dx_{2} = \frac{2}{L} (\frac{2L}{3\pi}) = \frac{4}{3\pi}$
$\frac{1}{L}\int_0^\frac{L}{2} sin^{2}(\frac{\pi x_{2}}{L})dx_{2} = \frac{1}{L} (\frac{L}{4}) = \frac{1}{4}$
$\frac{1}{L}\int_0^\frac{L}{2} sin^{2}(\frac{2\pi x_{1}}{L})dx_{1} = \frac{1}{L} (\frac{L}{4}) = \frac{1}{4}$

Putting this all together gives:

$\frac{2}{16}+\frac{16}{9\pi^{2}} = 0.31$

Hopefully I haven't made any silly mistakes...

I'm afraid you have.

 

[Zero1010]

Of course. Any hints where?

 

[PeroK]

Of course. Any hints where?

Look at post #13. Do your calculations agree with that?

 

[Zero1010]

No but the factor I am working with from the probability density is $\frac{1}{L}$ not $\frac{2}{L}$ in which is in post 13. Should I be using $\frac{2}{L}$?

 

[PeroK]

No but the factor I am working with from the probability density is $\frac{1}{L}$ not $\frac{2}{L}$ in which is in post 13. Should I be using $\frac{2}{L}$?

Here are the eigenfunctions:

$V_{A}(x) = \sqrt\frac{2}{L} sin (\frac{\pi x}{L})$

$V_{B}(x) = \sqrt\frac{2}{L} sin (\frac{2\pi x}{L})$

 

[PeroK]

PS It's not clear where you've gone wrong. You need to go through the calculation carefully.

 

[Zero1010]

Thanks for all your help and advice with this. Its gotten me so much closer to an answer than I would have been.

Time is becoming a bit of an issue as this is due in on Monday and I have one more question to look at. So I'm going to leave this for now and hopefully come back to it over the weekend and go through the calculations mare carefully.

Thanks again.

 

[PeroK]

Thanks for all your help and advice with this. Its gotten me so much closer to an answer than I would have been.

Time is becoming a bit of an issue as this is due in on Monday and I have one more question to look at. So I'm going to leave this for now and hopefully come back to it over the weekend and go through the calculations mare carefully.

Thanks again.

When you come back to this, you could follow the approach from post #13:
$$|\psi(x_1, x_2)|^2 = \frac 1 2 [\psi_1(x_1)^2 \psi_2(x_2)^2 + \psi_2(x_1)^2 \psi_1(x_2)^2 + 2\psi_1(x_1)\psi_2(x_1)\psi_1(x_2)\psi_2(x_2)]$$
$$\int_0^{\frac L 2}\psi_1(x)^2 dx = \frac 2 L \int_0^{\frac L 2}\sin^2(\frac{\pi x}{L}) dx = \frac 2 L (\frac L 4) = \frac 1 2$$
$$\int_0^{\frac L 2}\psi_2(x)^2 dx = \frac 2 L \int_0^{\frac L 2}\sin^2(\frac{2\pi x}{L}) dx = \frac 2 L (\frac L 4) = \frac 1 2$$
$$\int_0^{\frac L 2}\psi_1(x)\psi_2(x) dx = \frac 2 L \int_0^{\frac L 2}\sin(\frac{\pi x}{L})\sin(\frac{2\pi x}{L}) dx = \frac 2 L (\frac {2L}{3\pi}) = \frac 4 {3\pi}$$
And then it's just a case of plugging those numbers into the first equation.

 

[Zero1010]

Couldn't resist.

I understand the approach here and it is so much clearer than what I was doing before, the only thing I'm not clear on is the $\frac{2}{L}$ factor you have in the approach above.

When I worked out the spatial wave function at the start I multiplied the $\frac{\sqrt2}{\sqrt L}$ from the eigenfunctions with the $\frac{1}{\sqrt 2}$ from the spatial wavefunction equation to get $\frac{1}{\sqrt L}$ (Which looking back now I think should actually be $\frac{\sqrt2}{L}$ and so $\frac{2}{L^{2}}$ after squaring but this won't cancel out the L to get the probability so must be wrong)

Have i done this correctly or should I just be using the $\frac{\sqrt2}{\sqrt L}$ straight from the eigenfunction when I square to get the probability density which would give me the $\frac{2}{L}$ that you have above?

 

[PeroK]

Couldn't resist.

I understand the approach here and it is so much clearer than what I was doing before, the only thing I'm not clear on is the $\frac{2}{L}$ factor you have in the approach above.

When I worked out the spatial wave function at the start I multiplied the $\frac{\sqrt2}{\sqrt L}$ from the eigenfunctions with the $\frac{1}{\sqrt 2}$ from the spatial wavefunction equation to get $\frac{1}{\sqrt L}$ (Which looking back now I think should actually be $\frac{\sqrt2}{L}$ and so $\frac{2}{L^{2}}$ after squaring but this won't cancel out the L to get the probability so must be wrong)

Have i done this correctly or should I just be using the $\frac{\sqrt2}{\sqrt L}$ straight from the eigenfunction when I square to get the probability density which would give me the $\frac{2}{L}$ that you have above?

Yes, I know, but you didn't have an end-to-end calculation where I could see actually where you went wrong.

There's a problem with your answer of $\frac 1 8 + \frac{16}{9\pi^2}$ if you consider an antisymmetric wavefunction. The calculations are all the same but you would get $\frac 1 8 - \frac{16}{9\pi^2}$, which is less than zero.

You should have got $\frac 1 4 + \frac{16}{9\pi^2}$. This makes sense also because the $\frac 1 4$ is the classical answer, which you get without the cross terms. I was expecting an answer of $\frac 1 4 \pm X$ before I worked it out.

 

[PeroK]

Here's something else I noticed that helped me check my answer. We know that $\psi_1, \psi_2$ are orthonormal. So, we can write down immediately:
$$\int_{\frac L 2}^L \psi_1(x)^2 dx = \int_{\frac L 2}^L \psi_2(x)^2 dx = \frac 1 2$$
$$\int_{\frac L 2}^L\psi_1(x)\psi_2(x) dx = -\frac 4 {3\pi}$$
And we can see that the probability that the particles are found in the same half of the well is:
$$p(\text{same}) = \frac 1 2 + X = 2(\frac 1 4 + \frac X 2) $$
And, in different halves:
$$p(\text{different}) = \frac 1 2 - X = 2(\frac 1 4 - \frac X 2) $$
Where $X$ is some term involving $\pi^2$.

Note that I was able to see that by focussing on the properties of the eigenfunctions and leaving the normalisation constants alone!
[/QUOTE]

 

[Zero1010]

That all makes sense and good proof of the answer.

I have looked at the approach in #13 and #23 and I can now see where that factor came from.

thanks