Physical Meaning of QM Expectation Values and other ?s

[Rahmuss]

f95toli - Sounds like I need to get to work and make some discoveries that will simplify some things. :D

meopemuk - What I'm given is \Psi between one interval and everywhere else it is zero. A pretty simple infinite well problem. I asked my teacher about it and he explained that since \left\langle x\right\rangle was not dependent on time and \left\langle p\right\rangle was, then that's basically why I couldn't find it using \left\langle x\right\rangle.

reilly - My wife has a statistics book, I'll have to take a look. Yeah, I do need to get a handle on a lot of things in order to do well, and it goes back to the times when I didn't really understand the actual meaning of something, I was just memorizing the equations to get by.

 

[olgranpappy]

reilly - My wife has a statistics book, I'll have to take a look. Yeah, I do need to get a handle on a lot of things in order to do well, and it goes back to the times when I didn't really understand the actual meaning of something, I was just memorizing the equations to get by.

reilly's comment came off as a bit rude, to me at least--perhaps he didn't read all the posts previous to his.

I think that the book you really need to get is Griffiths' "Introduction to Quantum Mechanics". He reviews the necessary statistics in chapter one.

 

[meopemuk]

meopemuk - What I'm given is \Psi between one interval and everywhere else it is zero. A pretty simple infinite well problem. I asked my teacher about it and he explained that since \left\langle x\right\rangle was not dependent on time and \left\langle p\right\rangle was, then that's basically why I couldn't find it using \left\langle x\right\rangle.

I don't think your teacher's explanation is adequate. For "normal" Hamiltonians it is impossible to construct a time-dependent wave function for which \left\langle x\right\rangle is time-independent, and \left\langle p\right\rangle depends on time. However, there is no problem to do it the other way around: For example, in the case of a free particle wavepacket, the momentum is a conserved quantity, so \left\langle p\right\rangle = const, and the expectation value of position follows classical trajectory

\left\langle x(t)\right\rangle = \left\langle x(0)\right\rangle + \frac{1}{m} \langle p\rangle t

Actually, one can prove that expectation values of observables follow classical equations of motion. You can search for "Ehrenfest's theorem" in your QM textbook or on Google.

Eugene.

 

[Rahmuss]

[/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?

 

[meopemuk]

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

As far as I know, "coherent states" are just particular examples of wave packets. Classical formulas for expectation values are valid for a broader class of *reasonable* wave packets, for which average position and velocity can be well defined. You just need to consider wave packets described by a single "bump" in both the position and momentum spaces.

Eugene.

 

[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

<br /> \frac{2}{\pi}\int_0^\pi dx x^2 \sin^2(x) <br />

to

<br /> {\left(<br /> \frac{2}{\pi}\int_0^\pi dx x \sin^2(x)<br /> \right)}^2<br />

I find the correct inequality:

<br /> \frac{\pi^2}{3}&gt;\frac{\pi^2}{4}<br />

 

[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:

<br /> &lt;x^2&gt;=\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}<br />

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

<br /> &lt;x&gt;^2 = \frac{\pi^2}{4}<br />

So the inequality reads:

<br /> \frac{\pi^2}{3}-\frac{1}{2} &gt; \frac{\pi^2}{4}<br />

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

<br /> &lt;x&gt; = \frac{2}{\pi}(\frac{\pi^2}{4})=\frac{\pi}{2} <br />...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.