Harmonic oscillator - chance of of finding particle x>0

[renec112]

Homework Statement

A particle is moving in a 1-dimensional harmonic osciallator with the hamiltion:
$H = \hbar \omega (a_+ a_- + \frac{1}{2})$
at time $t=0$ the normalized wave function is given by
$\Psi(x,0) = \frac{1}{\sqrt{2}}(\psi_0(x) + i\psi_1(x))$
Task: Calculate for $t \geq 0$ the chance to meassure $x \geq 0$

2. Homework Equations
Well i think these equations are relevant. This is what i intend to use at least.
Chance of finding particle greater than 0:
$P(x\geq0) = \int_{0}^{\infty} \Psi^* \Psi dx$
Time dependent term for n'th psi:
$e^{-iE_n t / \hbar}$
The n'th psi can be written (Harmonic oscialtor):
$\psi_n(x) = (\frac{m \omega}{\pi \hbar})^{1/4} \frac{1}{\sqrt{2^n n!} } H_n(\xi) e^{-\xi^2 /2}$
where $H_n$ are hermite polynomials, and $\xi = \sqrt{\frac{m \omega}{\hbar}}x$ .

The Attempt at a Solution

Since it has to be for all times, i'll use the time dependent term and insert that on all the $\psi$'s.
Then, i'll just insert my values and integrate. However, i can't loose some energy parts... Let me show you (I'm not writing all constants in $e$ from the time dependent term):

$P(x\geq0) = \int_{0}^{\infty} \Psi^* \Psi dx = \int_{0}^{\infty} \frac{1}{\sqrt{2}}(\psi_0e^{iE_0} - i\psi_1e^{iE_1})\frac{1}{\sqrt{2}}(\psi_0e^{-iE_0} + i\psi_1e^{-iE_1}) dx$
$= \int_{0}^{\infty} \frac{1}{2}(\psi_0^2 + \psi_1^2 + i\psi_1\psi_0(e^{iE_0}e^{-E_1}-e^{iE_1}e^{-iE_0})) dx$
This is very i am stuck. I can't get rid of these annyoing time dependent terms. But if i could, i would try to insert the expression for $\psi_n$ given above, and try to integrate.

Any suggestions?

 

[TSny]

$P(x\geq0) = \int_{0}^{\infty} \Psi^* \Psi dx = \int_{0}^{\infty} \frac{1}{\sqrt{2}}(\psi_0e^{iE_0} - i\psi_1e^{iE_1})\frac{1}{\sqrt{2}}(\psi_0e^{-iE_0} + i\psi_1e^{-iE_1}) dx$
$= \int_{0}^{\infty} \frac{1}{2}(\psi_0^2 + \psi_1^2 + i\psi_1\psi_0(e^{iE_0}e^{-E_1}-e^{iE_1}e^{-iE_0})) dx$
This is very i am stuck. I can't get rid of these annyoing time dependent terms.

For convenience, I guess you left out $t$ and $\hbar$ in the exponentials. Note that $i(e^{iE_0}e^{-iE_1}-e^{iE_1}e^{-iE_0})$ can be written in a nice way in terms of a trig function.

Your integral contains three terms. Integrate each term separately. The time dependent part of the 3rd term can be factored outside the integral since it doesn't depend on $x$.

 

[renec112]

For convenience, I guess you left out $t$ and $\hbar$ in the exponentials. Note that $i(e^{iE_0}e^{-iE_1}-e^{iE_1}e^{-iE_0})$ can be written in a nice way in terms of a trig function.

Your integral contains three terms. Integrate each term separately. The time dependent part of the 3rd term can be factored outside the integral since it doesn't depend on $x$.

Thank you. Do you think i'ts the right approach? the calculations are getting a bit heavy for me..

 

[TSny]

Yes, right approach. The integrals from 0 to ∞ of ψ₀² and ψ₁² should not require any actual integrating if you think about it. The integral of ψ₀ψ₁ is easy to perform.

 

[PeroK]

Any suggestions?

I'd make a general suggestion, based on this and your post yesterday. I find it easier with these problems to generalise things - I find the structure of the algebra easier to see. Also, I find I learn the general algebraic patterns and can remember them better.

In this case, we have a wave function that is:

$\Psi(x, t) = \alpha \psi_0(x) \exp(-iE_0t/\hbar) + \beta \psi_1(x) \exp(-iE_1t/\hbar)$

Now, if you calculate $\Psi \Psi^*$, you should get:

$\Psi \Psi^* = |\alpha|^2\psi_0^2 + |\beta|^2\psi_1^2 + 2Re[\alpha \beta^* \exp(i(E_1 - E_0)t/\hbar] \psi_0 \psi_1$

This construction is the same whatever you are calculating and, as yesterday, is not specific to the Harmonic Oscillator. Then you substitute the specific wave functions and specific integrals.

I think it would pay dividends to work through that general calculation and try to remember that as a general algebraic approach.

 

[renec112]

I really appreciate both of you taking the time to help me out.

I'd make a general suggestion, based on this and your post yesterday. I find it easier with these problems to generalise things - I find the structure of the algebra easier to see. Also, I find I learn the general algebraic patterns and can remember them better.
I think it would pay dividends to work through that general calculation and try to remember that as a general algebraic approach.

I like that approach, i will try it. Looks cleaner and i see your point.I'm trying to solve the problem using your tips, but I'm not quite there yet, i think i have two problems. First:

The integrals from 0 to ∞ of ψ₀² and ψ₁² should not require any actual integrating if you think about it.

i'm not sure why $\psi_0^2$ and $\psi_1^2$ are easy. is it because they are orthonormal?

and my other problem:

Note that $i(e^{iE_0}e^{-iE_1}-e^{iE_1}e^{-iE_0})$ can be written in a nice way in terms of a trig function

did you wan't me to find $2Re[\alpha \beta^* \exp(i(E_1 - E_0)t/\hbar] \psi_0 \psi_1$ just like PeroK? I'm not sure how he got there.. I tried expanding using $e^{ix} = \cos(x) + i \sin(x)$ and $\sin(x) = (e^{ix}-e^{-ix}) / 2i$ but without luck.

I managed to do the integral of $\psi_1 \psi_0$, which i think is correct. Here it is
Using Hermite polynomials i can rewrite;
$\psi_0 = (\frac{m \omega}{\pi \hbar})^{1/4} \frac{1}{\sqrt{2}} e^{-\xi^2/2}$
$\psi_1 = (\frac{m \omega}{\pi \hbar})^{1/4} \frac{1}{\sqrt{2}} 2 \xi e^{-\xi^2/2}$
combine them in the integral:
$\int_{0}^{\infty} \psi_0 \psi_1 \ dx = \int_{0}^{\infty} (\frac{m \omega}{\pi \hbar})^{1/4} (\frac{m \omega}{\pi \hbar})^{1/4} \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} e^{-\xi^2/2} e^{-\xi^2/2} 2 \xi \ dx$
$= \int_{0}^{\infty}(\frac{m \omega}{\pi \hbar})^{2/4} \frac{1}{2} 2 e^{-\xi^2} \xi dx$
simplify further, insert $\xi$, and pull constant out of integral
$= \sqrt{\frac{m \omega}{\pi \hbar}} \int_{0}^{\infty} e^{-\xi^2} \xi dx$
$= \sqrt{\frac{m \omega}{\pi \hbar}} \int_{0}^{\infty} e^{-\frac{m\omega}{\hbar} x^2} \sqrt{\frac{m\omega}{\hbar}} x dx$
$=\frac{m \omega}{\sqrt{\pi} \hbar} \int_{0}^{\infty} e^{-\frac{m\omega}{\hbar} x^2} x dx$

i can solve that integral by using
$\int_{0}^{\infty} x^{2n+1} e^{-x^2/a^2} dx = \frac{n!}{2}a^{2n+2}$
with $n = 0$ and $a=-\sqrt{\frac{\hbar}{m \omega}}$

That gives me
$=\frac{m \omega}{\sqrt{\pi} \hbar} \int_{0}^{\infty} e^{-\frac{m\omega}{\hbar} x^2} x dx =\frac{m \omega}{\sqrt{\pi} \hbar} \frac{1}{2} \frac{\hbar}{m \omega} = \frac{1}{2\sqrt{\pi}}$

I hope that's correct. Then it's only the two last thing I'm missing for the task. Again, thank you so much for helping me.

 

[PeroK]

$=\frac{m \omega}{\sqrt{\pi} \hbar} \int_{0}^{\infty} e^{-\frac{m\omega}{\hbar} x^2} x dx =\frac{m \omega}{\sqrt{\pi} \hbar} \frac{1}{2} \frac{\hbar}{m \omega} = \frac{1}{2\sqrt{\pi}}$

I hope that's correct. Then it's only the two last thing I'm missing for the task. Again, thank you so much for helping me.

I got $\frac{1}{\sqrt{2\pi}}$

But, I did it with different variables, so it's hard to see where our solutions diverge.

You've got all the ideas, it's just the complexity of the integrals.

 

[PeroK]

$\psi_0 = (\frac{m \omega}{\pi \hbar})^{1/4} \frac{1}{\sqrt{2}} e^{-\xi^2/2}$

That $\sqrt2$ shouldn't be there. Which explains it.

 

[renec112]

That $\sqrt2$ shouldn't be there. Which explains it.

Oh yes, i see. off course.

Do i have the right idea about how ti simplify the time dependent term?

I see i don't have to do the $\psi_n^2$ term, because of course:
$\langle \psi_n | \psi_n \rangle = \int_\infty^\infty \psi_n * \psi_n dx = 1$
$\Rightarrow \int_0^\infty \psi_n * \psi_n dx = 1/2$

 

[PeroK]

Oh yes, i see. off course.

Do i have the right idea about how ti simplify the time dependent term?

I see i don't have to do the $\psi_n^2$ term, because of course:
$\langle \psi_n | \psi_n \rangle = \int_\infty^\infty \psi_n * \psi_n dx = 1$
$\Rightarrow \int_0^\infty \psi_n * \psi_n dx = 1/2$

Yes, you can get that by symmetry.

I don't see the problem with taking the real part of a complex exponential? You just have to be careful with the complex coefficient in there.

 

[renec112]

Yes, you can get that by symmetry.

I don't see the problem with taking the real part of a complex exponential? You just have to be careful with the complex coefficient in there.

I'm just wondering how you did it - how to simplify $i(e^{i \alpha}e^{- \beta}-e^{i \beta}e^{-i \alpha})$

 

[PeroK]

I'm just wondering how you did it - how to simplify $i(e^{i \alpha}e^{- \beta}-e^{i \beta}e^{-i \alpha})$

I'm on my phone, which can't cope with complex algebra. You need to combine the exponential products first. Then you could just write everything out: $exp = cos + isin$.

Better is to note that the second term is the conjugate of the first.

 

[PeroK]

Ps remember $z + z^* =2Re(z)$

 

[renec112]

I'm on my phone, which can't cope with complex algebra. You need to combine the exponential products first. Then you could just write everything out: $exp = cos + isin$.

Better is to note that the second term is the conjugate of the first.

Thanks for helping me on the go!

This is what i have tried. It's not the same but it looks reasonablei think. I went a bit crazy.
$i(e^{i \alpha}e^{-i \beta} - e^{i \beta}e^{- i \alpha})$
Drop the $i$ for now.
$e^{i \alpha}e^{-i \beta} - e^{i \beta}e^{- i \alpha} = e^{i (\alpha-\beta)} - e^{i (\beta - \alpha)}$
expand using euler.
$\cos(\alpha-\beta) + i \sin(\alpha-\beta) - \cos(\beta - \alpha) - \sin(\beta - \alpha) = \cos(\alpha-\beta) + i \sin(\alpha-\beta) - \cos(-1(\alpha-\beta)) - \sin(-1(\alpha- \beta))$
I can clean up, because:
$\sin(-A) = -\sin(A)$, and $\cos(-A) = \cos(A)$
giving me:
$\cos(\alpha-\beta) + i \sin(\alpha-\beta) - \cos(\alpha-\beta) + \sin(\alpha- \beta) = 2i \sin(\alpha-\beta) = -2i \sin(\beta-\alpha)$
Using the $i$ i saved later gives:
$2\sin(\beta-\alpha) = 2 \sin(\frac{t}{\hbar}(E_1 - E_0))$
I know the energies for a harmonic oscialtor ($E_n = (n + 1/2) \hbar \omega$).
$2\sin(\beta-\alpha) = 2 \sin(\frac{t}{\hbar}(\frac{3}{2} \hbar \omega - \frac{1}{2}\hbar \omega)) = 2 \sin(t \omega)$

Giving me the final change for finding the particle $x \geq 0$:
$1/2 (1/2 + 1/2 + 2 \sin(t \omega)) = 1/2 + 2 \sin(t \omega)$

I think it's ok since $t \omega$ is dimensionless..

 

[PeroK]

You forgot the term that came from integrating $\psi_0\psi_1$.

Otherwise, I think that's right.

 

[renec112]

You forgot the term that came from integrating $\psi_0\psi_1$.

Otherwise, I think that's right.

Ah yes i did.. Thanks for the help PeroK. :)