Evaluating A-Matrix Transformation for Hemitian Operator

[Bill Foster]

Homework Statement

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.

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

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:

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

 

[gabbagabbahey]

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

This is the transformation matrix.

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

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

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

What makes you think you are allowed to do this?

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

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

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

 

[jdwood983]

Homework Statement

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.

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

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:

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

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 $$f(A)$$:

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

 

[Bill Foster]

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

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."

What makes you think you are allowed to do this?

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

 

[Bill Foster]

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 $$f(A)$$:

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

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.

 

[jdwood983]

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.

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

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

 

[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:

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

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

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

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

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

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

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

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

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

 

[jdwood983]

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:

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

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

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

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

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

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

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

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

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

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.

 

[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.

 

[Bill Foster]

I guess those sums should be integrals.

 

[Bill Foster]

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

 

[Bill Foster]

Does this work here?

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

 

[Bill Foster]

If it does, then:

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

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

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

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

 

[jdwood983]

Does this work here?

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

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:

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

 

[jdwood983]

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.

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

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

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

 

[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:

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

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

 

[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.

 

[Bill Foster]

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

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

 

[Bill Foster]

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

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?

 

[Bill Foster]

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

 

[jdwood983]

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?

You seem to have forgotten that you have a $$\cos\theta$$ in your exponential, so you must include that in your integration over $$d(\cos\theta)$$. Note, also, that the problem suggests leaving the integral in some form for which you can't integrate any futher without numerical help (ie Mathematica, MATLAB, Maple, etc).

 

[gabbagabbahey]

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:

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

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

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

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

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

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

So far, so good

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

Are you sure about this?

 

[Bill Foster]

You seem to have forgotten that you have a $$\cos\theta$$ in your exponential, so you must include that in your integration over $$d(\cos\theta)$$. Note, also, that the problem suggests leaving the integral in some form for which you can't integrate any futher without numerical help (ie Mathematica, MATLAB, Maple, etc).

Yes, and wouldn't $$F(x')$$ also have some dependency on $$\phi$$ and $$\theta$$, since $$x'=r'=\sqrt{x'^2+y'^2+z'^2}$$?

 

[Bill Foster]

So far, so good



Are you sure about this?

Isn't...

$$\sum_k |a^{(k)}\rangle \langle a^{(k)}|=\textbf{1}$$

...the "identity" operator?

 

[gabbagabbahey]

Isn't...

$$\sum_k |a^{(k)}\rangle \langle a^{(k)}|=\textbf{1}$$

...the "identity" operator?

Sure, but what you've essentially done is say that

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

How can that possibly be?