Expectation Values: Prove It.

[ScotchDave]

The state $$\Psi$$ = $$\frac{1}{\sqrt{6}}$$$$\Psi$$_-1 + $$\frac{1}{\sqrt{2}}$$$$\Psi$$₁ + $$\frac{1}{\sqrt{3}}$$$$\Psi$$₂
is a linear combination of three orthonormal
eigenstates of the operator Ô corresponding
to eigenvalues -1, 1, and 2. What is the
expectation value of Ô for this state?
(A) 2/3
(B) $$\sqrt{\frac{7}{6}}$$
(C) 1
(D) 4/3
(E) $$\frac{\sqrt{3} + 2\sqrt{2} - 1}{\sqrt{6}}$$

$$<\hat{A}> = < \Psi |A|\Psi > = a < \Psi|\Psi > = a$$

So using this eqn I get <$$\hat{O}$$> = -1/6 + 1/2 + 2/3 = 1, this is the correct answer, but if someone could explain why this is correct I would appreciate it a lot.

 

[CompuChip]

The expectation value of $\hat O$ in the state $\Psi$ is
$$\langle \hat O \rangle_\Psi := \langle \Psi | \hat O | \Psi \rangle$$
as you said.
You can just write this out, using the linearity of the inner product and the orthonormality of the eigenfunctions. I'd write this out for you, but I want you to try it yourself first.

 

[ScotchDave]

Ok, so I wrote it out, but I still don't understand why

$$\left\langle$$$$\Psi$$|$$\hat{O}$$|$$\Psi$$$$\right\rangle$$ = o$$\left\langle$$$$\Psi$$|$$\Psi$$$$\right\rangle$$.

When I wrote it out I got: $$\left\langle$$$$\hat{O}$$$$\right\rangle$$ = o* $$\left\langle$$ ( $$\frac{1}{\sqrt{6}}$$$$\Psi$$_-1 + $$\frac{1}{\sqrt{2}}$$$$\Psi$$₁ + $$\frac{1}{\sqrt{3}}$$$$\Psi$$₂ ) * ( $$\frac{1}{\sqrt{6}}$$$$\Psi$$_-1 + $$\frac{1}{\sqrt{2}}$$$$\Psi$$₁ + $$\frac{1}{\sqrt{3}}$$$$\Psi$$₂ ) $$\right\rangle$$ = o ( $$\int$$ $$\frac{1}{\sqrt{6}}$$ * $$\frac{1}{\sqrt{6}}$$ * $$\Psi$$_-1 * $$\Psi$$_-1 dr + $$\int$$ $$\frac{1}{\sqrt{2}}$$ * $$\frac{1}{\sqrt{2}}$$ * $$\Psi$$₁ * $$\Psi$$₁ dr + $$\int$$ $$\frac{1}{\sqrt{3}}$$ * $$\frac{1}{\sqrt{3}}$$ * $$\Psi$$₂ * $$\Psi$$₂ dr ) =o (1/6 + 1/2 + 1/3) $$\neq$$ 1

Because $$\int$$ f(x)_n*f(x)_mdx = 0 if n$$\neq$$m and 1 if n=m. n=m in this case and also f(x)*=f(x) as no complex parts. I think...

I understand that the subscripts need to somehow be factors of their coefficients, but I don't know why that should be?

The forum won't let me preview the post, so I can't see if my equations are correct...

 

[badphysicist]

You should be getting a different o for each part of the wave function. For instance, o will equal -1 for one part, 1 for another and so on. In other words, o should not be out in front of everything at the end like you have in your equation since each psi has a different eigenvalue (o).

 

[ScotchDave]

So you're saying that:

$$\left\langle$$ $$\hat{O}$$ $$\right\rangle$$ = (-1*$$\int$$ $$\frac{1}{\sqrt{6}}$$ * $$\frac{1}{\sqrt{6}}$$ * $$\Psi$$_-1 * $$\Psi$$_-1) + (1*$$\int$$ $$\frac{1}{\sqrt{2}}$$ * $$\frac{1}{\sqrt{2}}$$ * $$\Psi$$₁ * $$\Psi$$₁) + (2*$$\int$$ $$\frac{1}{\sqrt{3}}$$ * $$\frac{1}{\sqrt{3}}$$ * $$\Psi$$₂ * $$\Psi$$₂)

Which due to the orthonormality of $$\Psi$$_-1, $$\Psi$$₁ and $$\Psi$$₂ is equal to:

-1/6+1/2+2/3, I think?

Even if this is correct guys, I still need to see a proof for the second step of: $$\left\langle$$ $$\hat{A}$$ $$\right\rangle$$ = $$\left\langle$$ $$\Psi$$ ¦A¦ $$\Psi$$$$\right\rangle$$ = a $$\left\langle$$$$\Psi$$¦$$\Psi$$$$\right\rangle$$

I've got a problem, I won't be able to remember this relationship unless I can see a derivation or a proof, guys please I need your help...

Thanks for the continuing help.

Dave

P.S. It still won't let me preview, but the checkers seem to think the LaTex is ok.

 

[CompuChip]

As an aside question, you seem to know quite a bit of LaTeX already. But why are you wrapping single characters in tex tags instead of just the whole formula?
So instead of

$$\langle$$$$\hat O$$$$\rangle$$ = $$\int$$$$\frac{1}{\sqrt{6}}$$ dx
just write
$$\langle \hat O \rangle = \int \frac{1}{\sqrt{6}} dx$$

That will also make it a lot more readable.

 

[ScotchDave]

Actually I know no LaTex at all, I'm using the toolbar on the forum, sorry that it's hard to read. You mean instead of writing (tex)\Psi(/tex)(tex)\Psi(/tex), I should write: (tex)\Psi\Psi(/tex)?

 

[triangleman]

Try evaluating <O> in a basis where O is diagonal. This is perfectly okay since <O> is independent of representation. Here's a hint: In the (-1, 1, 2)-basis, O will look like this

$$\left( \begin{array}{c c c} -1 & &\\ & 1 &\\ & & 2\\ \end{array} \right)$$

Just write psi as a vector, and you've got a straightforward multiplication problem: (psi)'* O (psi) = <O>

 

[ScotchDave]

Ok, so here's my working: $$\left\langle\hat{O}\right\rangle =\left\langle\Psi^{*}\left|\hat{O}\right|\Psi\right\rangle$$

$$=\left(\begin{array}{c}\frac{1}{\sqrt{6}}\Psi_{-1}\\\frac{1}{\sqrt{2}}\Psi_{1}\\\frac{1}{\sqrt{3}}\Psi_{2}\\\end{array}\right) \left(\begin{array}{c c c}-1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 2\\ \end{array}\right) \left( \frac{1}{\sqrt{6}}\Psi_{-1}, \frac{1}{\sqrt{2}}\Psi_{1}, \frac{1}{\sqrt{3}}\Psi_{2}\right)$$

$$=\left(\begin{array}{c}\frac{-1}{\sqrt{6}}\Psi_{-1}\\\frac{1}{\sqrt{2}}\Psi_{1}\\\frac{2}{\sqrt{3}}\Psi_{2}\\\end{array}\right)\cdot \left( \frac{1}{\sqrt{6}}\Psi_{-1}, \frac{1}{\sqrt{2}}\Psi_{1}, \frac{1}{\sqrt{3}}\Psi_{2}\right)$$

$$=\left( \frac{-1}{6}\Psi_{-1}\Psi_{-1} + \frac{1}{2}\Psi_{1}\Psi_{1} + \frac{2}{3}\Psi_{2}\Psi_{2}\right)$$

$$=\frac{-1}{6}+\frac{1}{2}+\frac{2}{3}=1$$
As the psi functions are orthonormal?

 

[CompuChip]

OK, so you have
$$\Psi = \frac{1}{\sqrt{6}} \Psi_{-1} + \frac{1}{\sqrt{2}} \Psi_1 + \frac{1}{\sqrt{3}} \Psi_2$$
where $\Psi_n$ are eigenfunctions of O, i.e.
(*) $$\hat O |\Psi_n\rangle = n \Psi_n$$ (n = -1, 1, 2)
By definition, the expectation value of O on this state is
$$\langle \Psi | \hat O | \Psi \rangle$$.
If you plug this in,
$$\langle \Psi | \hat O | \frac{1}{\sqrt{6}} \Psi_{-1} + \frac{1}{\sqrt{2}} \Psi_1 + \frac{1}{\sqrt{3}} \Psi_2 \rangle$$
Now the inner product is linear, so you are allowed to write this out as
$$\langle \Psi | \hat O | \frac{1}{\sqrt{6}} \Psi_{-1} \rangle + \langle \Psi | \hat O |\frac{1}{\sqrt{2}} \Psi_1 \rangle + \langle \Psi | \hat O |\frac{1}{\sqrt{3}} \Psi_2 \rangle$$
Scalars ("numbers") can be taken outside the whole thing, so you get
$$\frac{1}{\sqrt{6}} \langle \Psi | \hat O | \Psi_{-1} \rangle + \frac{1}{\sqrt{2}} \langle \Psi | \hat O |\Psi_1 \rangle + \frac{1}{\sqrt{3}} \langle \Psi | \hat O | \Psi_2 \rangle$$.
Now you can let O work on the wave functions and use that they are eigenfunctions, using the formula marked (*) above:
$$\frac{1}{\sqrt{6}} \langle \Psi | -1 \cdot \Psi_{-1} \rangle + \frac{1}{\sqrt{2}} \langle \Psi | 1 \cdot \Psi_1 \rangle + \frac{1}{\sqrt{3}} \langle \Psi | 2 \cdot \Psi_2 \rangle = \frac{-1}{\sqrt{6}} \langle \Psi | \Psi_{-1} \rangle + \frac{1}{\sqrt{2}} \langle \Psi | \Psi_1 \rangle + \frac{2}{\sqrt{3}} \langle \Psi | \Psi_2 \rangle$$
where I have taken the scalars out again.

Now let us look, for example, at the first term, $\frac{-1}{\sqrt{6}} \langle \Psi | \Psi_{-1} \rangle$. If we plug in the left side as well, this is
$$\frac{-1}{\sqrt{6}} \langle \frac{1}{\sqrt{6}} \Psi_{-1} + \frac{1}{\sqrt{2}} \Psi_1 + \frac{1}{\sqrt{3}} \Psi_2 | \Psi_{-1} \rangle$$
and since the inner product is also linear in the first slot you can again split this up and take out the scalars:
$$\frac{-1}{\sqrt{6}} \left( \frac{1}{\sqrt{6}} \langle \Psi_{-1} | \Psi_{-1} \rangle + \frac{1}{\sqrt{2}} \langle \Psi_1 | \Psi_{-1} \rangle + \frac{1}{\sqrt{3}} \langle \Psi_2 | \Psi_{-1} \rangle \right)$$
Because the eigenfunctions are orthonormal, $\langle \Psi_{-1} | \Psi_{-1} \rangle = 1$ and all the other inner products vanish, so you are left with
$$\frac{-1}{\sqrt{6}} \frac{1}{\sqrt{6}} 1 = - \frac{1}{6}$$.
Similarly, working this out for the other terms will give you 1/2 and 2/3.

 

[ScotchDave]

Thank you, suddenly it all makes sense. I'm not sure if I'm just too new to QM, or if I forgot the notation, but I think part of my problem with this question was notation. What do you use the vertical line to mean in this expression?

$$\frac{-1}{\sqrt{6}} \langle \Psi | \Psi_{-1} \rangle$$

By the way, is my LaTex better for you to read now?

Thanks Again

Dave

 

[CompuChip]

Probably a little of both. The concepts of QM, as well as its notation, take some time getting used to. Just keep doing it and eventually it'll be easier. I'm just using the same bra-ket notation as you did in your first post where you wrote $\langle \Psi | \Psi \rangle$, meaning the inner product (usually defined by $$\langle f | g \rangle = \int f^*(x) g(x) \, \mathrm dx$$). You could calculate it if the explicit form of Psi was given, but in this case all you need is that the inner product between two eigenfunctions yields 0 (if they are different) or 1 (if they are the same). Also you can use it to check that indeed
$$\langle a \Psi_1 + b \Psi_2 | \Psi_1 \rangle = a \langle \Psi_1 | \Psi_1 \rangle + b \langle \Psi_2 | \Psi_1 \rangle$$.

And yes, thank you, it's so much better to read now. Also note how all the strange errors have disappeared (they occurred for example because \left\langle and \right\rangle were not in the same formula, so TeX couldn't match them) and how everything is much better aligned vertically.

 

[ScotchDave]

Also, is there any way to put line breaks without closing the (tex) string?

 

[CompuChip]

You can using the align environment:
$$\begin{align*} a & = b \\ c &= d \\ 1 & \not= 0. \end{align*}$$
with & a column separator (the lines will be layed out per column) and \\ the line separator (click the formula to see the source, as I presume you found out by now).