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I got confused by the concept of the average value of an observable.

I know that when we measure a physical quantity $A$ in a specific state described by $\psi_1$, we only get the eigenvalue $a$ of the operator $\hat{A}$.

So theoretically the value of the physical quantity $A$ that we are measuring is equal to the eigenvalue of $\hat{A}\psi_1=a\psi_1$.

Then there is a formula to find the average value $\langle A\rangle$. And this is also an expectation value.


Does it mean that normally $\langle A\rangle=a$?


Then I get confused by the world ''average'' in ''average value $\langle A\rangle$''. Is this the average value of the physical quantity $A$ in all its the states?


And finally, are we assuming here that my system has only one system described by $\psi$? For example there is only one particle. Is it possible that I have three systems (three particles) described by different eigenfunction $\psi$ and eigenvalue $a$?

I am confusing everything.

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  • $\begingroup$ Typically when examining expectation values they are thought of as being statistical representations of the parameter being measured. We can obtain expectation values for $x$, $\hat{p}$, $E$, etc, by integrating across all allowed space, for the particle in question, the wave function, it's complex conjugate, and the operator itself. eg) $\langle x\rangle = \int\psi (x)\psi^*(x)x dx$ $\endgroup$ – bleuofblue Nov 15 '16 at 23:23
  • $\begingroup$ Further, $\psi$ intends to fully describe the state of a system. For three particles, three different wave functions would exist for the particles, which depend on the mass of the particle, and the potential which is bounding it. For three identical particles, the wave function and eigenstates will not differ, unless you consider time dependence in that the initial conditions of $\Psi(x,t=0)$ differ. $\endgroup$ – bleuofblue Nov 16 '16 at 0:00
  • $\begingroup$ Finally, if we consider the time independent schrodinger equation in that $\hat{\mathcal{H}}\psi = E\psi$, does it make sense that an energy eigenvalue $E$ = $\langle E \rangle$? They are clearly different quantities, as $E$ represents a quantized energy level, and $\langle E \rangle$ represents what we expect to measure for $E$ based on its whereabouts. This is the statistical representation I mentioned earlier. $\endgroup$ – bleuofblue Nov 16 '16 at 0:09
  • $\begingroup$ When you say ''For three identical particles, the wave function and eigenstates will not differ'', I understand why the wavefunction would be the same, but don't understand why the eigenstates would not differ. Could we also interpret that these three particules are in a periodic box? $\endgroup$ – Mariya Nov 16 '16 at 1:22
  • $\begingroup$ The periodic box is representative of the potential function in which the particle exists in. Consider the infinite square well, where particles are bounded by infinite potential walls. The time independent solutions for $\psi(x)$ will be the same for each particle, provided they are identical particles and there is nothing interacting with the particles aside from the potential function. The eigenstates will all be the same as well, unless each particle has different initial conditions from the other particles, which is defined by the state $\Psi(x,t)$ for $t=0$ $\endgroup$ – bleuofblue Nov 16 '16 at 1:32
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You are confusing a lot all right : )

Start with getting this equation $ \hat{A}|\psi_1 \rangle =a |\psi_1\rangle$ nailed down first.

You apply a Hermitian operator, $\hat{A}$ to a ket $|\psi_1 \rangle$ and it gives you an appropriate eigenvalue $a $.

So the $ \hat{X}$ operator gives you the position observable $x$, likewise applying $ \hat{P}$ gives you the real observable momentum $p $, associated with $|\psi_1 \rangle$.

I am assuming you know what using a Hermitian operator implies.

Does it mean that normally $<A>=a$?

You should read this website article later: Average Value for a fuller description but a summary for now is:

No, $<A>$ is not equal to $a$, in general.

$<A>$ is a prediction of the average value of repeated measurements, whereas $ \hat{A}|\psi_1 \rangle =a |\psi_1\rangle$ is the value of 1 (just 1) single measurement.

For example, say you have a wave function that you know describes the ground state of a particle $m $.

Now you want to find the average momentum $$<P> = \langle \psi_1 | \hat P|\psi_1\rangle$$

For our purposes, expectation value and average value are the same idea.

So when you work out $<P> = \langle \psi_1 | \hat P|\psi_1\rangle$ in the 1D ground state simple box potential, you are going to find that it's equal to 0.

That's because the average momentum of the particle in these conditions is zero. The particle has the same momentum in two opposite directions, so it's average value is zero.

Then I get confused by the world ''average'' in ''average value $\langle A\rangle$''. Is this the average value of the physical quantity $A$ in all its the states?

No.... It's the average value in one state, where you sandwich the A between, for example the functions for the ground state.

If you then want to find the average value for the first excited state, you sandwich the A between the functions that describe the first excited state.

You then keep doing this for any energy state that that you know the functions for.

enter image description here

Image Source: Energy Levels


And finally, are we assuming here that my system has only one system described by $\psi$? For example there is only one particle.

Yes, there is just 1 particle. No offence, but I would absolutely, definitely, completely stick to learning about one particle state for the moment.

Is it possible that I have three systems (three particles) described by different eigenfunction $\psi$ and eigenvalue $a$?

No. I would not worry at this time about 3 particle systems, that is normally further down the course, that is weeks or months way if you are just starting off.

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The average value of some observable is defined specifically as the average value of measurements on an ensemble of states. Think of it like a shelf and in that shelf you have 1,000,000 jars. In each jar you have the same state $\Psi$. Then you have 1,000,000 graduate students who each measure, say the $x$ position of all of these identical states. The average of all of these values will be your expectation value, $⟨x⟩$. So the expectation value is the average value of measurement on the same state, which is the crucial part.

A good example of this would be with a spin $\frac{1}{2}$ particle such as an electron. Say we have an electron in the $|+x⟩ = \frac{1}{\sqrt{2}} |+z⟩ + \frac{1}{\sqrt{2}} |-z⟩$ state, and we want to know the expectation value of the observable $S_z$ in this state. Well, you can compute it with the formula $$ ⟨S_z⟩ = \sum_{+,-} |c_n|^2 S_n,$$ or you can think about it conceptually. Again, you have 1,000,000 jars with this state in them. You get 1,000,000 PhD students now to make a measurement on the states, of the spin in the $z$ direction. You would expect that half of them be spin up ($+\frac{\hbar}{2}$) and have of them spin down ($-\frac{\hbar}{2}$), which we can also see from the coefficients of the $|+x⟩$ state. Thus, the expectation value $⟨S_z⟩ = 0$ in this case. I hope this clarifies your doubts.

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A quick clarification on average, mean and expectation value.

Here are two arrays of dice throws obtained with the following random generator:

$$\{x_1\}=(2,2,1,6,5,2,4,6,6,5)$$ $$\{x_2\}=(3,4,4,4,5,2,6,2,2,4,4,5,5,3,2,2,5,2,1,1)$$ The sample average (sample mean) of both is: $$\bar{x}_1=3.9$$ $$\bar{x}_2=3.25$$ Of course we know that the true mean (or population mean) is given by:

$$\langle\mu\rangle=\frac{1+2+3+4+5+6}{6}=3.5$$

This is what's known in quantum physics as the expectation value. But it's clear that this term is, in a sense at least, very much a misnomer: does anyone really expect to throw a $3.5$? Of course not: the probability of throwing a $3.5$ is $0$!

Similarly, the wave function of a particle in a 1D box of length $a$ is:

$$\psi_n(x)=\sqrt{\frac2a}\sin\Big(\frac{n\pi x}{a}\Big)\:\text{for } n=1,2,3,...$$

The expectation value of the position $\langle x\rangle$ can easily be calculated to be:

$$\langle x\rangle=\frac{a}{2}$$

But it can also be shown that the probability of finding the particle at $x=a/2$ is in fact $0$!

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protected by Qmechanic Nov 16 '16 at 7:22

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