Physical Meaning of QM Expectation Values and other ?s

[olgranpappy]

[/tex]olgranpappy - I actually do have Griffith's 'Introduction to Quantum Mechanics'. It's hard for me to read and understand (though I like his humor) because I'm more visual. When people start throwing equations in the mix it takes me 10 times longer to get through a page.

meopemuk - Are those the 'coherent' states that we're learning about now?

Also, I had a basic question on something. I'm trying to figure out expectation values better (how to get them). I want to get $$\left\langle x\right\rangle$$ and $$\left\langle x^{2}\right\rangle$$ for when $$\Psi (x) = sin(x)$$. And from my reading (and logically) $$\left\langle x\right\rangle^{2}$$ should be less than $$\left\langle x^{2}\right\rangle$$; but I'm not getting that. I'm choosing a region between 0 and $$\pi$$ (to make it easy). And I would expect (looking at a graph) that $$\left\langle x\right\rangle$$ would be $$\pi/2$$; but I'm getting something bigger than that. My normalization constent ends up being $$\sqrt{(2/\pi)}$$. And so my $$\left\langle x\right\rangle^{2}$$ is greater than my $$\left\langle x^{2}\right\rangle$$. What am I doing wrong?

are you sure you've got the right integrands... you should compare

$$\frac{2}{\pi}\int_0^\pi dx x^2 \sin^2(x)$$

to

$${\left( \frac{2}{\pi}\int_0^\pi dx x \sin^2(x) \right)}^2$$

I find the correct inequality:

$$\frac{\pi^2}{3}>\frac{\pi^2}{4}$$

 

[Rahmuss]

meopemuk - We're just learning about coherent states. Our teacher said they are also known as Glauber states. He said one of the first major questions they had about QM was where it seems to give the same results as classical physics. And Glauber in 1963 (according to Wikipedia) found them. They are states which minimize uncertainty (ie. $$\sigma_{x}\sigma_{p}$$ exactly equals $$\hbar /2$$. When I saw your comment about classical trajectory I thought maybe that was one of those cases; but again we have only talked about them for one day so far.

olgranpappy - Wait, you're using $$A^{2}$$ instead of just $$A$$, yet in all my notes it's showing just $$A$$. But ignoring my normalization constant I do indeed get $$\frac{\pi^{2}}{4}$$ for $$\left\langle x\right\rangle$$ (but that's before squaring it). Let me try $$\left\langle x^{2}\right\rangle$$ again... using the indefinite integral form:

$$\int x^{2}sin^{2}(ax)ds = \frac{x^{3}}{6} - sin(2ax)(\frac{x^{2}}{4a} - \frac{1}{8a^{3}}) - \frac{xcos(2ax)}{4a^{2}}$$

And letting $$a=1$$. Not using my normalization constant ($$A = \sqrt{2/\pi}$$). I get

$$\left\langle x^{2}\right\rangle = \frac{\pi^{3}}{6}-\frac{\pi}{4}$$
Because when you evaluate it at $$x=0$$ the whole side disappears (we have $$xcos(2x)$$ instead of just $$cos(2x)$$ like we did with $$\left\langle x\right\rangle$$, so you don't have anything to cancel the extra term as you did when you evaluated it as $$x=\pi$$.

So after squaring $$\left\langle x\right\rangle$$ I get the values:

$$\left\langle x\right\rangle^{2} = \frac{\pi^{4}}{16} = 6.088$$
$$\left\langle x^{2}\right\rangle = \frac{\pi^{3}}{6} - \frac{\pi}{4} = 4.382$$

Maple agrees with me. So what am I missing?

 

[olgranpappy]

meopemuk - We're just learning about coherent states. Our teacher said they are also known as Glauber states. He said one of the first major questions they had about QM was where it seems to give the same results as classical physics. And Glauber in 1963 (according to Wikipedia) found them. They are states which minimize uncertainty (ie. $$\sigma_{x}\sigma_{p}$$ exactly equals $$\hbar /2$$. When I saw your comment about classical trajectory I thought maybe that was one of those cases; but again we have only talked about them for one day so far.

olgranpappy - Wait, you're using $$A^{2}$$ instead of just $$A$$, yet in all my notes it's showing just $$A$$. But ignoring my normalization constant I do indeed get $$\frac{\pi^{2}}{4}$$ for $$\left\langle x\right\rangle$$ (but that's before squaring it). Let me try $$\left\langle x^{2}\right\rangle$$ again... using the indefinite integral form:

$$\int x^{2}sin^{2}(ax)ds = \frac{x^{3}}{6} - sin(2ax)(\frac{x^{2}}{4a} - \frac{1}{8a^{3}}) - \frac{xcos(2ax)}{4a^{2}}$$

And letting $$a=1$$. Not using my normalization constant ($$A = \sqrt{2/\pi}$$). I get

$$\left\langle x^{2}\right\rangle = \frac{\pi^{3}}{6}-\frac{\pi}{4}$$
Because when you evaluate it at $$x=0$$ the whole side disappears (we have $$xcos(2x)$$ instead of just $$cos(2x)$$ like we did with $$\left\langle x\right\rangle$$, so you don't have anything to cancel the extra term as you did when you evaluated it as $$x=\pi$$.

So after squaring $$\left\langle x\right\rangle$$ I get the values:

$$\left\langle x\right\rangle^{2} = \frac{\pi^{4}}{16} = 6.088$$
$$\left\langle x^{2}\right\rangle = \frac{\pi^{3}}{6} - \frac{\pi}{4} = 4.382$$

Maple agrees with me. So what am I missing?

Oops, sorry, about that. I should have written:

$$<x^2>=\frac{2}{\pi}\int_0^\pi dx x^2 \sin^2(x) = \frac{2}{\pi}\left(\frac{\pi^3}{6}-\frac{\pi}{4}\right)=\frac{\pi^2}{3}-\frac{1}{2}$$

but, the previous evaluation of <x>^2 is okay. I.e., still:

$$<x>^2 = \frac{\pi^2}{4}$$

So the inequality reads:

$$\frac{\pi^2}{3}-\frac{1}{2} > \frac{\pi^2}{4}$$

I.e., 2.79 > 2.47...it looks like you used your normalization in one case but not the other. you messed up on <x>. It's

$$<x> = \frac{2}{\pi}(\frac{\pi^2}{4})=\frac{\pi}{2}$$...oh actually, you didn't use your normalization in either case. I think you need to do that because the proof that <x^2> > <x>^2 relies on the fact that <1> = 1

 

[meopemuk]

meopemuk - We're just learning about coherent states. Our teacher said they are also known as Glauber states. He said one of the first major questions they had about QM was where it seems to give the same results as classical physics. And Glauber in 1963 (according to Wikipedia) found them. They are states which minimize uncertainty (ie. $$\sigma_{x}\sigma_{p}$$ exactly equals $$\hbar /2$$. When I saw your comment about classical trajectory I thought maybe that was one of those cases; but again we have only talked about them for one day so far.

I don't think that classical limit can be obtained only for special Glauber states. The center of any *reasonable* wave packet will move along classical trajectory.

Eugene.

 

[Rahmuss]

olgranpappy - Oh shoot you're right. I need to apply my normalization constant BEFORE I square $$\left\langle x\right\rangle$$. Oops... ok, let me check it again...

Wow, what do you know. It helps when I can do math. :D

$$\left\langle x\right\rangle^{2} = \frac{\pi^{2}}{4} = 2.467$$

$$\left\langle x^{2}\right\rangle = \frac{\pi^{3}}{6} - \frac{\pi}{4} = 2.79$$

 

[Rahmuss]

So, I thought that $$\left\langle x^{2}\right\rangle$$ would always be greater than $$\left\langle x\right\rangle^{2}$$. Yet, when I setup a graph in Maple of $$\sigma_{x}$$ using $$sin(x)$$ or $$cos(x)$$ it shows a fun little curve that drops out (goes negative) when the range for the test is negative (obviously); but curiously enough it also drops to negative around 4.4 for $$sin(x)$$ and around 3 for $$cos(x)$$. I first noticed it on paper when I was choosing the range from $$0 to 2\pi$$. I kept getting a negative number inside the square root for determining $$\sigma_{x}$$. So then I set it up in Maple with $$sin(x)$$ and got good values all the way up until I tried $$0 to 2\pi$$ again. Then I decided to graph it and found that $$\sigma_{x}$$ is imaginary when using a range greater than $$2\pi$$.

So, what's wrong with this picture? Try calculating $$\left\langle x\right\rangle^{2}$$ and $$\left\langle x^{2}\right\rangle$$ using $$sin(x)$$ with the range from $$0 to 2\pi$$.

What am I missing? This really throws a kink into what I'm trying to research (not that I have the knowledge to research anything seriously) to get a better understanding of things.