# Physical Meaning of QM Expectation Values and other ?s

1. Sep 2, 2007

### 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.

2. Sep 2, 2007

### 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).

3. Sep 2, 2007

### meopemuk

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.

4. Sep 2, 2007

### malawi_glenn

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

5. Sep 2, 2007

### meopemuk

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.

6. Sep 2, 2007

### olgranpappy

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.

Last edited: Sep 2, 2007
7. Sep 2, 2007

### meopemuk

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.

8. Sep 3, 2007

### 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).

Last edited: Sep 3, 2007
9. Sep 3, 2007

### olgranpappy

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.

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.

10. Sep 3, 2007

### meopemuk

$$x^{2}$$ 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.

11. Sep 3, 2007

### 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 $$x$$?

12. Sep 3, 2007

### meopemuk

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

Eugene.

13. Sep 3, 2007

### Rahmuss

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.

14. Sep 3, 2007

### cesiumfrog

15. Sep 3, 2007

### 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.

16. Sep 3, 2007

### meopemuk

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

$$\langle x^2 \rangle \neq \langle x \rangle^2 [/itex] Does it answer your question? Eugene. 17. Sep 4, 2007 ### cesiumfrog Keep thinking about air molecules. The kinetic energy of a molecule is proportional to the square of it's momentum. Since the temperature is not absolute zero, most-every molecule has positive kinetic energy, and so an average <p^2> is strictly positive. But if there is no wind (equal numbers of molecules moving left as moving right) the average *net* momentum is exactly zero. 18. Sep 4, 2007 ### Rahmuss meopemuk and cesiumfrog - Yes, that makes more sense. I was thinking [tex]\left\langle x^{2}\right\rangle = \left\langle x\right\rangle ^{2}$$; 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.

19. Sep 4, 2007

### cesiumfrog

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.

20. Sep 4, 2007

### 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?