Physical Meaning of QM Expectation Values and other ?s

[Rahmuss]

I am just starting an introduction to quantum mechanics this semester, and it's hard for me to do some of my homework and follow some of the lectures because I can't grasp the actual 'physical' meaning of some of the concepts.

What do they mean by the expectation values? For example $$\left\langle x \right\rangle$$ and $$\left\langle x^{2} \right\rangle$$ and $$\left\langle p \right\rangle$$ and so on? I know the basic formula. I know what the operators look like; but I don't know what they mean physically. Does the expectation value of $$\left\langle x \right\rangle$$ mean the position value (along the x-axis) where they would expect to find the particle? And is that the same for the momentum $$\left\langle p \right\rangle$$?

I have more questions; but I'll have to post them later. Any thoughts would be great. Thanks.

 

[cesiumfrog]

Yeah, eg. if they repeated the experiment and measured position each time, the average (er.. "mean") result should match the expectation value.

Note that doesn't mean you will ever measure that exact value. In air (classically), you might say <p>=0 (no wind) but <p^2> > 0 (finite temperature).

 

[meopemuk]

What do they mean by the expectation values? For example $$\left\langle x \right\rangle$$ and $$\left\langle x^{2} \right\rangle$$ and $$\left\langle p \right\rangle$$ and so on? I know the basic formula. I know what the operators look like; but I don't know what they mean physically. Does the expectation value of $$\left\langle x \right\rangle$$ mean the position value (along the x-axis) where they would expect to find the particle? And is that the same for the momentum $$\left\langle p \right\rangle$$?

In the quantum world, if you prepare N identical copies of the particle (all of them are in the same state) and measure their positions, there is no guarantee that you'll obtain N identical results. Most often (if particle state is not an eigenstate of the position operator) you will obtain a statistical distribution of x values. The expectation value $$\left\langle x \right\rangle$$ is just the average measured position, or the center of the statistical distribution.

The same can be said about any other particle observable (momentum, energy, spin projection, etc.)

Eugene.

 

[malawi_glenn]

Also, it should be mentioned that it is the same in ordinary mathematical statistics. Mean, expectation value, standard deviation, dispersion, variance etc.

 

[meopemuk]

Also, it should be mentioned that it is the same in ordinary mathematical statistics. Mean, expectation value, standard deviation, dispersion, variance etc.

Yes, I just wanted to add that in classical statistical physics the variance of measurements is always produced by the variance of preparation conditions and, with careful preparation of states, the variance can be reduced to zero. In quantum mechanics, the variance is present even if all members of the ensemble are prepared in the same state. This is the most important difference between classical and quantum physics.

Eugene.

 

[olgranpappy]

In quantum mechanics, the variance is present even if all members of the ensemble are prepared in the same state.

An exception being, of course, if you can manage to prepare the system in an eigenstate of the operator in question.

For example, if I prepare the system in an eigenstate of the Hamiltonian with energy E, then I will always find

$$<H>=E$$

$$<H^2>=E^2$$

etc... And in particular

$$<(H-<H>)^2>=0$$

I.e., no variance.

Similarly if I could prepare the system in an eigenstate of the position operator (with eigenvalue $$x_0$$, say) I would always find

$$<x>=x_0$$

and

$$<(x-<x>)^2>=0$$
...of course, a real physical system can't actually be an eigenstate of the position operator... but you get the picture.

 

[meopemuk]

An exception being, of course, if you can manage to prepare the system in an eigenstate of the operator in question.

Yes, this is true. If the ensemble is prepared in an eigenstate of an operator F, then measurements of this observable will not have variance. However measurements of other observables (whose operators do not commute with F) will have a variance.

Eugene.

 

[Rahmuss]

Wow, thank you all for your comments. They help a lot. Here are a couple of other questions.

cesiumfrog - So, in an actual experiment you would have three spatial dimensions $$(x, y, z)$$ and make measurements for where you found the particle, and the more measurements you make the more likely the mean value will be at the expectation value?

meopemuk - What about $$x^{2}$$? What is the physical meaning of $$x^{2}$$?

olgranpappy - What's an eigenstate? I mean, they've presented the concept in class; but I do not fully understand it. They use too much math to explain it. I need more of a visual picture to understand it. What does $$\left\langle\left(H-\left\langle H\right\rangle\right)^{2}\right\rangle=0$$ mean? What's the difference between $$H$$ and $$\left\langle H\right\rangle$$? Apparently $$\left\langle H\right\rangle$$ is an operator defined as $$-\left(\hbar /2m) * (\partial^{2}/\partial x^{2}) + V(x)$$, so I'm not looking for a mathematical definition, I'm asking for you to describe what $$\left\langle H\right\rangle$$ and $$H$$ are without using mathematical notation (ie. as if you were describing it to someone and you didn't have anything to write on).

 

[olgranpappy]

What's an eigenstate?

It's analogous to an "eigenvector" of a matrix (in quantum mechanics, operators can be represented by matrices and states can be represented by vectors).

Operators "act on" states by transforming them into different states and in general the state you "get out" is not the same state you "put in". If the state that you get out is proportional to the state you put in then that state is called an eigenstate.

Let me make an analogy with an operation that can be performed in real space (easier to visualize that abstract state-space most times). Consider a z-axis in real space to point upwards and then consider the operation of "rotation about the z-axis". So, now consider the two empty beer bottles I have on my table right now: Bottle A is standing up, and Bottle B is laying on its side. If I "operate" on the bottle A it still looks the same (because it is cylindrically symmetric and I rotate it about the z-axis). The operation didn't do anything, so it is an eigenbottle of the operation "rotation about z-axis". On the other hand, if I rotate bottle B then it will generally look different, so it is not an eigenbottle.

I mean, they've presented the concept in class; but I do not fully understand it. They use too much math to explain it. I need more of a visual picture to understand it. What does $$\left\langle\left(H-\left\langle H\right\rangle\right)^{2}\right\rangle=0$$ mean? What's the difference between $$H$$ and $$\left\langle H\right\rangle$$? Apparently $$\left\langle H\right\rangle$$ is an operator defined as $$-\left(\hbar /2m) * (\partial^{2}/\partial x^{2}) + V(x)$$, so I'm not looking for a mathematical definition, I'm asking for you to describe what $$\left\langle H\right\rangle$$ and $$H$$ are without using mathematical notation (ie. as if you were describing it to someone and you didn't have anything to write on).

As explained a bit above, operators "act on" states and transform them into different states. In this case, H is an operator called the Hamiltonian which is analogous to the total energy in classical mechanics. The symbol <H> means that I let H act on a state (here denoted by nothing inside a bracket "|>" but often called "Psi" as in "|psi>") and
after acting it on a state I get out a different state and then I take the inner product of the state I got out with the state I put in:

H|psi> = |chi>

<psi|chi> = expectation value of Hamiltonian.

But, really, the expectation value of an operator and its interpretation is a fundamental of quantum mechanics: The expectation value is a real number and it gives the "expected" value of a measurement of the energy. I.e., if I have N identical states and I measure the energy of each of them then add the energies up and divide by N I get approximately <H>.

Note that if |psi> is an eigenstate of H then

H|psi> = E|psi>

where E is just a number so that <psi|H|psi>=E<psi|psi>=E

Similarly, H^2 is an operator, a operator which means "operate with H twice" and thus
<H^2> is the expected value of the operator H^2.

Also, (H-<H>) is an operator. Really, since <H> is a number, it means the operator
(H-<H>I) where I is the identity operator (the operator which "does nothing").

I|psi>=|psi> for and psi.

So too is (H-<H>)^2 an operator. It means that you apply the above operator two times. And <(H-<H>)^2> is its expected value.

 

[meopemuk]

meopemuk - What about $$x^{2}$$? What is the physical meaning of $$x^{2}$$?

$$x^{2}$$ is just a formal function of the position operator. In order to measure values of this observable, you simply measure $$x$$ and take squares of the measured values.

If you want to have a "physical" realization of this observable, imagine that markings on your rulers have quadratic dependence on the distance from the origin (rather than the usual linear dependence).

In the same fashion you can define any other function of $$x$$ or of any other observable. Similarly, there is no difficulty to define (multivariable) functions of any set of mutually commuting observables.

Eugene.

 

[Rahmuss]

olgranpappy - Wow, great explanations. Ok, I'm beginning to see the 'why' in some of the things I'm learning and it's great. Good example for me to visualize what you mean by eigenstates. It's in a state which does not change, or if it changes, then it's simply a proportional change (like increasing by a simple multiplicity). If $$\left\langle H\right\rangle$$ is in a different state than $$H$$, then why when we subtract $$\left\langle H\right\rangle$$ from $$H$$ and square it, do we get zero (ie. $$\left\langle \left( H - \left\langle H\right\rangle \right) ^{2} \right\rangle=0$$)? I've seen a couple of different mathematical equations representing $$\left\langle H\right\rangle$$. One where they use a single partial derivative and one where they use the second partial derivative, which one is right?

meopemuk - Thank you for that explanation. So then why do we use $$x^{2}$$ instead of just using $$x$$?

 

[meopemuk]

meopemuk - So then why do we use $$x^{2}$$ instead of just using $$x$$?

I don't understand your question. Could you give an example where $$x^{2}$$ is used instead of $$x$$?

Eugene.

 

[Rahmuss]

You had said:

$$x^{2}$$ is just a formal function of the position operator. In order to measure values of this observable, you simply measure $$x$$ and take squares of the measured values.

So it sounded like you would get just as usable information finding $$x$$ as you would $$x^{2}$$. So I was just wondering why they even have the $$x^{2}$$ operator. Or maybe I'm not understanding what you initially said about it.

In our homework we are told to find the expectation value of $$x$$; but we are also told to find the expectation value of $$x^{2}$$, so I was just wondering why we do that if the one is just the square of the other. So apparently I'm missing something here.

 

[cesiumfrog]

In our homework we are told to find the expectation value of $$x$$; but we are also told to find the expectation value of $$x^{2}$$, so I was just wondering why we do that if the one is just the square of the other. So apparently I'm missing something here.

Read what I wrote above.

 

[Rahmuss]

cesiumfrog - It seems like if <p> = 0, then <p^2> would also = 0. Because from what I've read in this section is that the square of the expectation value is simply the square of the result (ie. the found expectation value), and so if the found expectation value for momentum is zero, then it's square (ie. 0^2) should also be zero. I guess that's what I'm missing, I'm not seeing what you are referring to, how it's not zero.

 

[meopemuk]

In our homework we are told to find the expectation value of $$x$$; but we are also told to find the expectation value of $$x^{2}$$, so I was just wondering why we do that if the one is just the square of the other. So apparently I'm missing something here.

The average (or expectation) value of a random variable $$x^{2}$$ is generally not equal to the square of the average value of $$x$$

[tex] \langle x^2 \rangle \neq \langle x \rangle^2 [/itex]

Does it answer your question?

Eugene.

 

[cesiumfrog]

Keep thinking about air molecules. The kinetic energy of a molecule is proportional to the square of it's momentum. Since the temperature is not absolute zero, most-every molecule has positive kinetic energy, and so an average <p^2> is strictly positive. But if there is no wind (equal numbers of molecules moving left as moving right) the average *net* momentum is exactly zero.

 

[Rahmuss]

meopemuk and cesiumfrog - Yes, that makes more sense. I was thinking $$\left\langle x^{2}\right\rangle = \left\langle x\right\rangle ^{2}$$; but with the example given by cesiumfrog I think I can see how that isn't the case. We're not talking about simple values; but about larger functions.

 

[cesiumfrog]

We're not talking about simple values; but about larger functions.

I worry I'm giving the wrong impression there. Write down a list of 15 random "simple values". Compute the square of the mean. Then compute the mean of the square.

 

[Rahmuss]

Randomized - 2, 8, 4, 6, 7, 2, 8, 1, 9, 0, 3, 4, 2, 6, 7
Ordered - 0, 1, 2, 2, 2, 3, 4, 4, 6, 6, 7, 7, 8, 8, 9

mean = 4, average = 4.6

square of mean = 16

------------------------
squared = 0, 1, 4, 4, 4, 9, 16, 16, 36, 36, 49, 49, 64, 64, 81

mean of the square = 16

Is this what you mean? Or is this my impression of what you mean? The numbers are what I mean by "simple values". Though what we are dealing with in Quantum Mechanics, is not simple values, it's functions right?

 

[Son Goku]

Similarly if I could prepare the system in an eigenstate of the position operator (with eigenvalue $$x_0$$, say) I would always find

$$<x>=x_0$$

and

$$<(x-<x>)^2>=0$$
...of course, a real physical system can't actually be an eigenstate of the position operator... but you get the picture.

Just a small correction, generally systems that begin in the eigenstate of some operator don't remain in that eigenstate, but evolve into a more general state. It's usually only the eigenstates of the Hamiltonian that remain the same always. Except for systems like the Hydrogen atom where you have states which are both angular momentum and energy eigenstates.

 

[meopemuk]

Randomized - 2, 8, 4, 6, 7, 2, 8, 1, 9, 0, 3, 4, 2, 6, 7
Ordered - 0, 1, 2, 2, 2, 3, 4, 4, 6, 6, 7, 7, 8, 8, 9

mean = 4, average = 4.6

square of mean = 16

------------------------
squared = 0, 1, 4, 4, 4, 9, 16, 16, 36, 36, 49, 49, 64, 64, 81

mean of the square = 16

Is this what you mean? Or is this my impression of what you mean? The numbers are what I mean by "simple values". Though what we are dealing with in Quantum Mechanics, is not simple values, it's functions right?

Your numbers are a good example of a possible series of measurements of the quantum observable x. In your case the expectation value of x is $\langle x \rangle = 4.6$, and the expectation value of $x^2$ is $\langle x^2 \rangle = 28.9$. Then the relationship we were talking about $\langle x \rangle^2 \neq \langle x^2 \rangle$ becomes obvious as $4.6^2 \neq 28.9$.

The quantity that you call "mean" (I think, its other name is "median") is of no particular interest for quantum mechanics.

Eugene.

 

[olgranpappy]

cesiumfrog - It seems like if <p> = 0, then <p^2> would also = 0. Because from what I've read in this section is that the square of the expectation value is simply the square of the result (ie. the found expectation value), and so if the found expectation value for momentum is zero, then it's square (ie. 0^2) should also be zero. I guess that's what I'm missing, I'm not seeing what you are referring to, how it's not zero.

No, the point is that

$$<p^2>\neq<p>^2$$

in general... only if |> is an eigenstate of the p operator.

 

[olgranpappy]

If $$\left\langle H\right\rangle$$ is in a different state than $$H$$,

Sorry, but what you wrote above doesn't make sense;

"<H>" depends on what the state "|>" is, and "<H>" is is a number. Whereas, "H" (not sandwiched between two states) is an operator...

I hate to say it, but math might just be unavoidable in trying to understand what these quantities mean. I mean, sure there is a physical interpretation, but you must really get used to the formalism if you want to ever solve real (or textbook) problems using QM.

Also, as a general rule, it is very good to speak with your teacher about these things. Asking him (or her) similar question to those you asked us will let him know you are interested in learning, and if you are in a class can only help your grade. There is no substitute for a good conversation with a good teacher.

Cheers,

Adam

 

[jtbell]

Randomized - 2, 8, 4, 6, 7, 2, 8, 1, 9, 0, 3, 4, 2, 6, 7
Ordered - 0, 1, 2, 2, 2, 3, 4, 4, 6, 6, 7, 7, 8, 8, 9

mean = 4, average = 4.6

What you call the "mean" is actually the "median." "Mean" and "average" are the same thing. As Eugene pointed out, we don't deal with the median in QM. The expectation value of a random variable is the expected mean of a large number of samples of that variable. More precisely, it is the limit of the mean as the number of samples goes to infinity.

The mean of a finite number of samples is itself a random variable!

 

[Rahmuss]

meopemuk and jtbell - Ah, yes, thanks for clarifying that up for me. I was thinking of median instead of the average.

olgranpappy - Yes, I do indeed need to learn the math behind it, it's just easier for me to learn what the physical meaning of it is, and then I understand the math behind it more intuitively. Then I can actually grasp other concepts before they've even been presented because my mind has already questioned and answered them. If I don't understand the physical meaning, then my mind just seems to have a really hard time remembering anything.

I have talked with my teacher via email; but I can't use the same symbols in email and so it's hard to explain what I mean when I don't understand the physical meaning behind it. His office hours don't work with my schedule (working full time and only getting about 4 hours of sleep at night), and his TA's office hours don't work either, so I've scheduled a special time one day a week for 30 minutes with his TA, which is helpful; but he tends to show me all the easy stuff rather than explaining what I don't truly understand.

There is also a problem on my homework (and this might belong in the homework section) where it says to find $$\left\langle p\right\rangle$$ and it says that we cannot find it using $$md\left\langle x\right\rangle /dt$$; but it doesn't say why. So I'm wondering why. It might have to do with the value for $$\left\langle x\right\rangle$$ though I'm not sure. It looks like $$\left\langle x\right\rangle$$ may not exist (or that it is zero), is that possible?

 

[f95toli]

olgranpappy - Yes, I do indeed need to learn the math behind it, it's just easier for me to learn what the physical meaning of it is, and then I understand the math behind it more intuitively. Then I can actually grasp other concepts before they've even been presented because my mind has already questioned and answered them. If I don't understand the physical meaning, then my mind just seems to have a really hard time remembering anything.

I understand what you are saying. However, the problem is that many of the things we calculate in QM do NOT have a "physical meaning" in the conventional meaning of the word; even if it is something you can (at least in principle) measure it is so different from anything you come across in your everyday life that the ONLY way to understand it is via math (spin would be a good example, even something as "simple" as a Stern-Gerlach experiment is quite strange when you think about it).

Once you get used to QM (no one ever truly understand QM, it is too weird, we just get used to it) you will start thinking about the "real world" in terms of math and "physical meaning" becomes less important.

 

[olgranpappy]

There is also a problem on my homework (and this might belong in the homework section) where it says to find $$\left\langle p\right\rangle$$ and it says that we cannot find it using $$md\left\langle x\right\rangle /dt$$; but it doesn't say why.

The velocity operator times the mass is not always the same as the momentum operator, for example, this would be the case if the potential depends on momentum (e.g. interaction with external vector potential).

What are you given? Wavefunctions in position space? Remember, the representation of the momentum operator $$\hat p$$ is always given by $$-i\frac{d}{dx}$$ in position space.

 

[meopemuk]

The velocity operator times the mass is not always the same as the momentum operator, for example, this would be the case if the potential depends on momentum (e.g. interaction with external vector potential).

This is correct. However, in the case of non-relativistic potential, which does not depend on momentum, there should be no problem to write

$$\langle \hat{p} \rangle = m\frac{d}{dt} \langle \hat{x}(t) \rangle |_{t=0}$$

Eugene.

 

[reilly]

Rahmuss -- Get a book on statistics. Things like means and expectation values, and variances are very basic. Of course (<x>)*2 does not equal <x*x>, for virtually all probability distributions -- check out the formula for variances. Get that statistics book -- or material from Google. You really need to master this stuff in order to do well in QM.

Frankly, I'm amazed at how many people in this thread do not understand basic statistics, nor elementary QM.

Regards,
Reilly Atkinsonj

 

[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 dependant 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 dependant 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.