# Sakurai 1.27

1. Oct 26, 2009

### Bill Foster

1. The problem statement, all variables and given/known data

Suppose that $$f\left(A\right)$$ is a function of a Hemitian operator A with the property $$A|a'\rangle=a'|a'\rangle$$. Evaluate $$\langle b''|f\left(A\right)|b'\rangle$$ when the transformation matrix from the a' basis to the b' basis is known.

$$\langle\beta|A|\alpha\rangle=\int dx' \int dx'' \langle\beta|x'\rangle\langle x'|A|x''\rangle \langle x''|\alpha\rangle=\int dx' \int dx'' \psi_\beta^*\left(x'\right)\langle x'|A|x''\rangle\psi_\alpha\left(x''\right)$$

3. The attempt at a solution

Transformation from $$a'$$ to $$b'$$ basis:

$$U|a'\rangle = |b'\rangle$$

Multiply both sides by $$a''$$:

$$\langle a''|U|a'\rangle = \langle a''|b'\rangle$$

This is the transformation matrix.

Now insert $$|a''\rangle\langle a''|$$ into $$\langle b''|f\left(A\right)|b'\rangle$$:

$$\langle b''|f\left(A\right)|a''\rangle\langle a''|b'\rangle$$

Now insert $$|a'\rangle\langle a'|$$ into the above:

$$\langle b''|a'\rangle\langle a'|f\left(A\right)|a''\rangle\langle a''|b'\rangle$$

This is a transformation matrix: $$\hat{T}=\langle a''|b'\rangle$$

And so is this: $$\hat{T}^*=\langle b''|a'\rangle$$.

So then:

$$\langle b''|f\left(A\right)|b'\rangle = \hat{T}^* \langle a'|f\left(A\right)|a''\rangle \hat{T}$$

Last edited: Oct 26, 2009
2. Oct 26, 2009

### gabbagabbahey

No it isn't. In fact, it isn't even a matrix; just an inner product between two states.

What makes you think you are allowed to do this?

Again, not matrices. Also, $\langle a''|b'\rangle^{\dagger}=\langle b'|a''\rangle\neq\langle b''|a'\rangle$

3. Oct 26, 2009

### jdwood983

If you have $$\langle b''|f\left(A\right)|b'\rangle$$, why not just insert $$\sum_{a'}|a'\rangle\langle a'|$$ in there so you can get the eigenvalue of $$f(A)$$:

$$\sum_{a'}\langle b''|f(A)|a'\rangle\langle a'|b'\rangle=\sum_{a'}f(a')\langle b''|a'\rangle\langle a'|b'\rangle$$

Last edited: Oct 26, 2009
4. Oct 26, 2009

### Bill Foster

Ahhh...I looked at Sakurai a little closer. You're right.

Section 1.5:

"The matrix elements of the U operator are built up of the inner products of old base bras and new base kets."

If Chuck Norris can divide by zero, then I can do that.

5. Oct 26, 2009

### Bill Foster

If

$$A|a'\rangle=a'|a'\rangle$$

Then

$$f\left(A\right)|a'\rangle=f\left(a'\right)|a'\rangle$$ ??

I was wondering how that eigenvalue relationship came into play in this problem.

Thanks.

6. Oct 26, 2009

### jdwood983

Yeah, you can prove this by Taylor expanding some function and then applying the ket, for example the exponential function:

$$\begin{array}{ll}\exp[mA]|a'\rangle&=\left(1+mA+\frac{m^2A^2}{2!}+\cdots\right)|a'\rangle \\ \,&=|a'\rangle+mA|a'\rangle+\frac{m^2}{2!}A^2|a'\rangle+\cdots \\ \,&=|a'\rangle+ma'|a'\rangle+\frac{m^2}{2!}(a')^2|a'\rangle+\cdots \\ \,&=\left(1+ma'+\frac{(ma')^2}{2!}+\cdots\right)|a'\rangle \\ \,&=\exp[ma']|a'\rangle$$

7. Oct 26, 2009

### Bill Foster

So here's what I came up with. Please let me know how it looks:

$$\langle b^{(m)}|f\left(A\right)|b^{(n)}\rangle$$

Insert the following:

$$\sum_k |a^{(k)}\rangle \langle a^{(k)}|$$
$$\sum_l |a^{(l)}\rangle \langle a^{(l)}|$$

$$\sum_k \sum_l \langle b^{(m)}|a^{(k)}\rangle \langle a^{(k)}|f\left(A\right)|a^{(l)}\rangle \langle a^{(l)}|b^{(n)}\rangle$$

$$=\sum_k \sum_l f\left(a^{(l)}\right) \langle b^{(m)}|a^{(k)}\rangle \langle a^{(k)}|a^{(l)}\rangle \langle a^{(l)}|b^{(n)}\rangle$$

$$=\sum_k \sum_l f\left(a^{(l)}\right) \langle b^{(m)}|a^{(k)}\rangle \delta_{kl} \langle a^{(l)}|b^{(n)}\rangle$$

$$=\sum_k f\left(a^{(k)}\right) \langle b^{(m)}|a^{(k)}\rangle \langle a^{(k)}|b^{(n)}\rangle$$

$$=\sum_k f\left(a^{(k)}\right) \langle a^{(m)}|U^{\dagger}|a^{(k)}\rangle \langle a^{(k)}|U|a^{(n)}\rangle$$

$$=f\left(a^{(k)}\right) \langle a^{(m)}|U^{\dagger}U|a^{(n)}\rangle$$

$$=f\left(a^{(k)}\right) \langle a^{(m)}|a^{(n)}\rangle$$

$$=f\left(a^{(k)}\right) \delta_{nm}$$

8. Oct 26, 2009

### jdwood983

It looks like you don't actually need to add $$|a^{(k)}\rangle\langle a^{(k)}|$$ in the above, but otherwise seems fine to me.

9. Oct 26, 2009

### Bill Foster

Any ideas on how to do part (b)?

I'm approaching it the same way, with some differences:

$$\langle \textbf{p}''|F\left(r\right)|\textbf{p}'\rangle$$

$$= \sum_{\textbf{x}'} \sum_{\textbf{x}''}\langle \textbf{p}''|\textbf{x}''\rangle \langle \textbf{x}'' |F\left(r\right)|\textbf{x}'\rangle \langle \textbf{x}'|\textbf{p}'\rangle$$

$$\langle \textbf{x}'|\textbf{p}' \rangle = \frac{1}{\left(2\pi\hbar\right)^{\frac{3}{2}}}e^{i \frac{\textbf{p}'\cdot\textbf{x}'}{\hbar}}$$

$$= \sum_{\textbf{x}'} \sum_{\textbf{x}''}\frac{1}{\left(2\pi\hbar\right)^{\frac{3}{2}}}e^{-i \frac{\textbf{p}''\cdot\textbf{x}''}{\hbar}} \langle \textbf{x}'' |F\left(r\right)|\textbf{x}'\rangle \frac{1}{\left(2\pi\hbar\right)^{\frac{3}{2}}}e^{i \frac{\textbf{p}'\cdot\textbf{x}'}{\hbar}}$$

$$= \frac{1}{\left(2\pi\hbar\right)^{3}} \sum_{\textbf{x}'} \sum_{\textbf{x}''}e^{-i \frac{\textbf{p}''\cdot\textbf{x}''}{\hbar}} \langle \textbf{x}'' |F\left(r\right)|\textbf{x}'\rangle e^{i \frac{\textbf{p}'\cdot\textbf{x}'}{\hbar}}$$

Now I'm stuck.

10. Oct 26, 2009

### Bill Foster

I guess those sums should be integrals.

11. Oct 26, 2009

### Bill Foster

$$\frac{1}{\left(2\pi\hbar\right)^{3}} \int \int e^{-i \frac{\textbf{p}''\cdot\textbf{x}''}{\hbar}} \langle \textbf{x}'' |F\left(r\right)|\textbf{x}'\rangle e^{i \frac{\textbf{p}'\cdot\textbf{x}'}{\hbar}}d^3x' d^3x''$$

12. Oct 26, 2009

### Bill Foster

Does this work here?

$$F(r)|\textbf{x}'\rangle = F(x')|\textbf{x}'\rangle$$

13. Oct 26, 2009

### Bill Foster

If it does, then:

$$\frac{1}{\left(2\pi\hbar\right)^{3}} \int \int F\left(x'\right)e^{-i \frac{\textbf{p}''\cdot\textbf{x}''}{\hbar}} \langle \textbf{x}'' |\textbf{x}'\rangle e^{i \frac{\textbf{p}'\cdot\textbf{x}'}{\hbar}}d^3x' d^3x''$$

$$=\frac{1}{\left(2\pi\hbar\right)^{3}} \int \int F\left(x'\right)e^{-i \frac{\textbf{p}''\cdot\textbf{x}''}{\hbar}} \delta^3\left(\textbf{x}''-\textbf{x}'\right) e^{i \frac{\textbf{p}'\cdot\textbf{x}'}{\hbar}}d^3x' d^3x''$$

$$=\frac{1}{\left(2\pi\hbar\right)^{3}} \int F\left(x'\right)e^{-i \frac{\textbf{p}''\cdot\textbf{x}'}{\hbar}} e^{i \frac{\textbf{p}'\cdot\textbf{x}'}{\hbar}}d^3x'$$

$$=\frac{1}{\left(2\pi\hbar\right)^{3}} \int F\left(x'\right)e^{-i \frac{\textbf{x}'}{\hbar}\cdot\left(\textbf{p}''-\textbf{p}'\right)}d^3x'$$

14. Oct 26, 2009

### jdwood983

while $$r=\sqrt{x^2+y^2+z^2}$$, you can't make this claim. You'l have to solve this integral as a function of $$F(\mathbf{r})$$. Also, an integral is also a sum so

$$\sum_{a'}|a'\rangle\langle a'|=\int|a'\rangle\langle a'|\,da'=1$$

Basically your last post has it mostly right, though you can't use $$F(x')$$. Expanding the integral for spherical coordinates:

$$\langle \mathbf{p}''|F(\mathbf{r})|\mathbf{p}'\rangle=\frac{1}{\left(2\pi\hbar\right)^3}\int_0^{2\pi}d\phi'\int_{-1}^1d(\cos\theta)\int_0^\infty (r')^2F(r')\exp\left[\frac{i|\mathbf{p}'-\mathbf{p}''|\cdot\mathbf{r}}{\hbar}\right]\,dr'$$

15. Oct 26, 2009

### jdwood983

As before, I don't think you need the double sum. You should just have

$$\begin{array}{ll}\langle \textbf{p}''|F\left(\mathbf{r}\right)|\textbf{p}'\rangle&= \sum_{\textbf{r}'} \langle \textbf{p}''|F\left(\mathbf{r}\right)|\textbf{r}'\rangle \langle \textbf{r}'|\textbf{p}'\rangle \\ &=\sum_{r'}F(\mathbf{r}')\langle\mathbf{p}''|\mathbf{r}'\rangle\langle\mathbf{r}'|\mathbf{p}'\rangle\end{array}$$

note the use of $$\mathbf{r}$$ instead of $$\mathbf{x}$$ because $$F$$ is a function of $$r$$, not $$x$$.

Last edited: Oct 26, 2009
16. Oct 26, 2009

### jdwood983

Actually, this last line should read

$$\langle \mathbf{p}''|F(\mathbf{r})|\mathbf{p}'\rangle=\frac{1}{\left(2\pi\hbar\right)^3}\int_0^{2\pi}d\phi\int_{-1}^1d(\cos\theta)\int_0^\infty (r')^2F(r')\exp\left[\frac{i|\mathbf{p}'-\mathbf{p}''|r'\cos\theta}{\hbar}\right]\,dr'$$

I just forgot to remove the vector off of $$\mathbf{r}'$$, but is otherwise correct as is

17. Oct 26, 2009

### jdwood983

Yeah...I made a mistake on this post and since we can't delete them, I'm just erasing it all. You should be fine with what I had written in posts 15 and 16.

Last edited: Oct 26, 2009
18. Oct 27, 2009

### Bill Foster

When I say $$\textbf{x}$$, I mean $$\textbf{x}=x\hat{\textbf{x}}+y\hat{\textbf{y}}+z\hat{\textbf{z}}$$

And $$\textbf{x}'=x'\hat{\textbf{x}}+y'\hat{\textbf{y}}+z'\hat{\textbf{z}}$$

19. Oct 27, 2009

### Bill Foster

So evaluate this:

$$\langle\mathbf{p}''|F(\mathbf{r})|\mathbf{p}'\rangle=\frac{1}{\left(2\pi\hbar\right)^3}\int_0^{2\pi}d\phi\int_{-1}^1d(\cos\theta)\int_0^\infty (r')^2F(r')\exp\left[\frac{i|\mathbf{p}'-\mathbf{p}'|r'\cos\theta}{\hbar}\right]\,dr'$$

And get this:

$$\langle\mathbf{p}''|F(\mathbf{r})|\mathbf{p}'\rangle=\frac{4\pi}{\left(2\pi\hbar\right)^3} \int_0^\infty (r')^2F(r')\exp\left[\frac{i|\mathbf{p}'-\mathbf{p}'|r'\cos\theta}{\hbar}\right]\,dr'$$

What next? integrate by parts?

20. Oct 27, 2009

### Bill Foster

Looks like integrating by parts will get rid of the exponential leaving me something in terms of $$F(r')$$.