Expectation Value of f(x): Physical Meaning

[Technon]

The expectation value of any function $f(x)$ is given by <f(x)>= \int_{-\infty}^{\infty}f(x)\psi^2(x) dx
But what is $f(x)$ actually? In a physical sense.

For example if $f(x)=x$ or $f(x)=x^2$, what do these functions represent on a physical level?

 

[PeterDonis]

The expectation value of any function $f(x)$ is given by

In QM, expectation values aren't properties of functions; they're properties of operators. The expectation value of an operator $\hat{A}$ is (note that we're using a particular representation, the one-dimensional position representation, in all of this--things look different in other representations)

$$
\langle \hat{A} \rangle = \int_{- \infty}^{\infty} \psi^* (x) \hat{A} \psi (x) dx
$$

For example, the position operator $\hat{Q}$ is just $x$, i.e., "multiply by $x$"; so its expectation value is

$$
\langle \hat{Q} \rangle = \int_{- \infty}^{\infty} x \psi^* (x) \psi (x) dx
$$

which is similar to what you wrote (note that we can rearrange the factors inside the integral because multiplication of functions is commutative). Similarly, an operator that was something like "multiply by $f(x)$", where $f(x)$ is some function of $x$, would have an expectation value that looks similar to what you wrote.

However, this won't be true for all operators, because not all operators can be represented as "multiply by some function $f(x)$". For example, the momentum operator $\hat{P}$ is $- i \hbar d / dx$, so its expectation value is

$$
\langle \hat{P} \rangle = - i \hbar \int_{- \infty}^{\infty} \psi^* (x) \frac{d \psi (x)}{dx} dx
$$

which is not similar to what you wrote.

what do these functions represent on a physical level?

As should be evident from the above, the functions represent operators, but, as noted, not all operators can be represented as functions.

 

[Technon]

I'm sure that's very good but I really just want to know some examples of what the functions represent physically.

Example: Find (a) $<x>$ and (b) $<x^2>$ for a particle in its ground state in a box of length L.

So I calculated these and got some values, and I checked the answers are correct, but what did I actually calculate? I mean what do these functions and their expectation value represent on a physical (not just mathematical) level?

 

[PeterDonis]

I'm sure that's very good but I really just want to know some examples of what the functions represent physically.

And I answered that: they represent operators (though not all operators can be so represented). I even told you explicitly which operator "multiply by $x$" represents.

Example

Where is this example from?

 

[PeroK]

I'm sure that's very good but I really just want to know some examples of what the functions represent physically.

Example: Find (a) $<x>$ and (b) $<x^2>$ for a particle in its ground state in a box of length L.

So I calculated these and got some values, and I checked the answers are correct, but what did I actually calculate? I mean what do these functions and their expectation value represent on a physical (not just mathematical) level?

Those are the expected values of a) a measurement of position and b) the square of a measurement of position.

 

[Technon]

Those are the expected values of a) a measurement of position and b) the square of a measurement of position.

Thanks for a direct answer. Measurement of position makes sense, because the answer was L/2, which means the expected value for the position is in the center of the box which makes sense.

"Square of a measurement of position" however, seems more like a mathematical abstraction than representing something physical. Or does it also represent some physical quality, like for example velocity, acceleration, potential energy, etc?

Where is this example from?

Physics for Scientists and Engineers, 6th Edition

 

[PeroK]

"Square of a measurement of position" however, seems more like a mathematical abstraction than representing something physical. Or does it also represent some physical quality, like for example velocity, acceleration, potential energy, etc?

It's essentially the potential energy in a harmonic oscillator: $\frac12 m \omega^2 x^2$.

In general, you could take $f(x)$ to be the function of position determined by your potential.

 

[Technon]

It's essentially the potential energy in a harmonic oscillator: $\frac12 m \omega^2 x^2$.

In general, you could take $f(x)$ to be the function of position determined by your potential.

Ok. The answer for that question was 0.283L^2. Does that mean that this is the expected value for the potential energy of the particle in the box?

 

[PeroK]

Ok. The answer for that question was 0.283L^2. Does that mean that this is the expected value for the potential energy of the particle in the box?

How would I know that? If you have specific problems, you ought to post them in homework.

 

[Technon]

How would I know that? If you have specific problems, you ought to post them in homework.

The question is not whether that answer is correct, which I already verified it is. The question is what the answer means. Does <x^2> represent the expected value for the potential energy of the particle in the box?

 

[PeroK]

The question is not whether that answer is correct, which I already verified it is. The question is what the answer means. Does <x^2> represent the expected value for the potential energy of the particle in the box?

The particle in a box usually means it has no potential energy. It's not a harmonic oscillator, in any case.

But, $\langle x^2 \rangle$ is also needed to calculate the variance and standard deviation for position measurements.

 

[Technon]

The particle in a box usually means it has no potential energy.

<x^2> represents expected value of the potential energy and it has been calculated to be not zero, then that seems incorrect.

 

[PeterDonis]

<x^2> represents expected value of the potential energy

No, $\frac{1}{2} m \omega^2 <x^2>$ represents the expected value of the potential energy for a harmonic oscillator. As @PeroK has already pointed out, a particle in a box is not a harmonic oscillator.

As I've already asked, where did you get this question from? It would help if we knew the context.

[Edited to fix user reference above.]

 

[Technon]

No, $\frac{1}{2} m \omega^2 <x^2>$ represents the expected value of the potential energy for a harmonic oscillator. As @Ibix has already pointed out, a particle in a box is not a harmonic oscillator.

So what does only <x^2> represent?

As I've already asked, where did you get this question from? It would help if we knew the context.

I already answered, see post above. Unfortunately there isn't more information given in the example written there than what I've written here.

 

[PeterDonis]

So what does only <x^2> represent?

Multiplying or dividing by numbers doesn't change the physical meaning, just the units. If you pick units such that $\frac{1}{2} m \omega^2 = 1$, then the expectation value of potential energy for a harmonic oscillator is just $<x^2>$.

 

[Ibix]

As @Ibix has already pointed out, a particle in a box is not a harmonic oscillator.

@PeroK, I presume. "Refresh knowledge of QM" is still on my to-do list.

 

[PeterDonis]

I already answered, see post above.

Ah, got it.

Physics for Scientists and Engineers, 6th Edition

Looking through what I can see online in this textbook, it appears to be taking a very rote learning approach--tell you some basic stuff, tell you how to calculate things, but not explain the physical models very much. I don't see anything at all about operators in the QM section (of course not all of it is visible online and I don't have the actual book), and to me that's a key concept in QM since it's one of the main concepts that doesn't occur in classical mechanics, and since it's central to a fundamental change in the nature of models between classical mechanics and QM.

In any case, for the particle in a box, which is what it appears to be asking for expectation values for, the operator $x^2$ doesn't have any physical meaning apart from being the square of the position operator, and useful in computing averages and variances. So I'm not sure why the textbook is even asking for its expectation value to be computed, since it doesn't talk about any of that stuff that I can see.

 

[PeterDonis]

@PeroK, I presume.

Oops, yes. Will edit my post to fix.

 

[Nugatory]

"Square of a measurement of position" however, seems more like a mathematical abstraction than representing something physical. Or does it also represent some physical quality, like for example velocity, acceleration, potential energy, etc?

Informally, it tells us how tightly localized the particle is. If you measure the position of the particle many times and average the results, you will get something close to the expectation value of $x$; but if the expectation value of $\langle{x}^2\rangle$ is large compared with $\langle{x}\rangle^2$ then it is likely that many of the individual measurements will be far from $\langle{x}\rangle$. That's part of what @PeroK was getting at in the second sentence of post #11 above.
(Do be aware that to make what I just wrote work properly always you have to label your x-axis in such a way that $\langle{x}\rangle=0$. This is always possible, and doing so allows my simplified explanation of the physical significance of $\langle{x}^2\rangle$).

As other have pointed out, $\langle{x}^2\rangle$ is also the potential energy of a harmonic oscillator. That's because of the specific properties of the harmonic oscillator potential so is not generally true for all quantum systems... But if you think about how the potential energy of a harmonic oscillator varies with distance from the center, it will be intuitively obvious that $\langle{x}^2\rangle$ should have this physical meaning in this particular system.

 

[A. Neumaier]

$<f(x)>$ is approximately the value of the average of $f(x)$ when $x$ runs over independent measurements of the observable $x$ (here the position). $\sqrt{<f(x)^2>-<f(x)>^2}$ is approximately the value of the corresponding standard deviation, giving the intrinsic uncertainty of these measurements. In particular, $<x^2>$ is needed to calculate the uncertainty of the position measurements.

 

[Technon]

Looking through what I can see online in this textbook, it appears to be taking a very rote learning approach--tell you some basic stuff, tell you how to calculate things, but not explain the physical models very much. I don't see anything at all about operators in the QM section (of course not all of it is visible online and I don't have the actual book), and to me that's a key concept in QM since it's one of the main concepts that doesn't occur in classical mechanics, and since it's central to a fundamental change in the nature of models between classical mechanics and QM.

In any case, for the particle in a box, which is what it appears to be asking for expectation values for, the operator $x^2$ doesn't have any physical meaning apart from being the square of the position operator, and useful in computing averages and variances. So I'm not sure why the textbook is even asking for its expectation value to be computed, since it doesn't talk about any of that stuff that I can see.

I absolutely agree, I'm disappointed in some sections of the book are only mentioned briefly without proper explanations and they would have needed more information.

I made some pictures of the related parts from the book, if it helps: