Physical Meaning of QM Expectation Values and other ?sby Rahmuss Tags: expectation, meaning, physical, values 

#1
Sep207, 01:07 AM

P: 223

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 [tex]\left\langle x \right\rangle[/tex] and [tex]\left\langle x^{2} \right\rangle[/tex] and [tex]\left\langle p \right\rangle[/tex] 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 [tex]\left\langle x \right\rangle[/tex] mean the position value (along the xaxis) where they would expect to find the particle? And is that the same for the momentum [tex]\left\langle p \right\rangle[/tex]? I have more questions; but I'll have to post them later. Any thoughts would be great. Thanks. 



#2
Sep207, 01:33 AM

P: 2,050

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
Sep207, 01:42 AM

P: 1,743

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



#4
Sep207, 02:46 AM

Sci Advisor
HW Helper
P: 4,739

Physical Meaning of QM Expectation Values and other ?s
Also, it should be mentioned that it is the same in ordinary mathematical statistics. Mean, expectation value, standard deviation, dispersion, variance etc.




#5
Sep207, 03:57 AM

P: 1,743

Eugene. 



#6
Sep207, 04:09 AM

HW Helper
P: 1,273

For example, if I prepare the system in an eigenstate of the Hamiltonian with energy E, then I will always find [tex] <H>=E [/tex] [tex] <H^2>=E^2 [/tex] etc... And in particular [tex] <(H<H>)^2>=0 [/tex] I.e., no variance. Similarly if I could prepare the system in an eigenstate of the position operator (with eigenvalue [tex]x_0[/tex], say) I would always find [tex] <x>=x_0 [/tex] and [tex] <(x<x>)^2>=0 [/tex] ...of course, a real physical system can't actually be an eigenstate of the position operator... but you get the picture. 



#7
Sep207, 01:44 PM

P: 1,743

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
Sep307, 03:13 PM

P: 223

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 [tex](x, y, z)[/tex] 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 [tex]x^{2}[/tex]? What is the physical meaning of [tex]x^{2}[/tex]? 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 [tex]\left\langle\left(H\left\langle H\right\rangle\right)^{2}\right\rangle=0[/tex] mean? What's the difference between [tex]H[/tex] and [tex]\left\langle H\right\rangle[/tex]? Apparently [tex]\left\langle H\right\rangle[/tex] is an operator defined as [tex]\left(\hbar /2m) * (\partial^{2}/\partial x^{2}) + V(x)[/tex], so I'm not looking for a mathematical definition, I'm asking for you to describe what [tex]\left\langle H\right\rangle[/tex] and [tex]H[/tex] are without using mathematical notation (ie. as if you were describing it to someone and you didn't have anything to write on). 



#9
Sep307, 03:56 PM

HW Helper
P: 1,273

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 statespace most times). Consider a zaxis in real space to point upwards and then consider the operation of "rotation about the zaxis". 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 zaxis). The operation didn't do anything, so it is an eigenbottle of the operation "rotation about zaxis". On the other hand, if I rotate bottle B then it will generally look different, so it is not an eigenbottle. 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: Hpsi> = chi> <psichi> = 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 Hpsi> = Epsi> where E is just a number so that <psiHpsi>=E<psipsi>=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"). Ipsi>=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
Sep307, 03:59 PM

P: 1,743

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 [tex]x[/tex] or of any other observable. Similarly, there is no difficulty to define (multivariable) functions of any set of mutually commuting observables. Eugene. 



#11
Sep307, 09:08 PM

P: 223

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 [tex]\left\langle H\right\rangle[/tex] is in a different state than [tex]H[/tex], then why when we subtract [tex]\left\langle H\right\rangle[/tex] from [tex]H[/tex] and square it, do we get zero (ie. [tex]\left\langle \left( H  \left\langle H\right\rangle \right) ^{2} \right\rangle=0[/tex])? I've seen a couple of different mathematical equations representing [tex]\left\langle H\right\rangle[/tex]. 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 [tex]x^{2}[/tex] instead of just using [tex]x[/tex]? 



#12
Sep307, 09:19 PM

P: 1,743

Eugene. 



#13
Sep307, 10:18 PM

P: 223

You had said:
In our homework we are told to find the expectation value of [tex]x[/tex]; but we are also told to find the expectation value of [tex]x^{2}[/tex], 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
Sep307, 10:33 PM

P: 2,050





#15
Sep307, 11:10 PM

P: 223

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
Sep307, 11:32 PM

P: 1,743

[tex] \langle x^2 \rangle \neq \langle x \rangle^2 [/itex] Does it answer your question? Eugene. 



#17
Sep407, 01:11 AM

P: 2,050

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, mostevery 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
Sep407, 01:20 AM

P: 223

meopemuk and cesiumfrog  Yes, that makes more sense. I was thinking [tex]\left\langle x^{2}\right\rangle = \left\langle x\right\rangle ^{2}[/tex]; 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.



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