Product of an Antisymmetric Tensor and a Symmetric Tensor

A well-known homework problem from General Relativity and Tensor Analysis textbooks asks the student to prove that:

$A^{\alpha \beta }B_{\alpha \beta } = 0$,

if A is an antisymmetric $\left ( \frac{2}{0} \right )$ tensor and B is a symmetric $\left ( \frac{0}{2} \right )$ tensor.

Now, this problem will do your head in if you treat the above equation as a matrix product. Remember that we're looking at Einstein Summation Notation here and all will be fine. If we sum on repeated indices, we can work things out in the following way:

$A^{\alpha \beta }B_{\alpha \beta } =-A^{\beta\alpha}B_{\beta\alpha}$ (antisymmetry of A and symmetry of B)     (1)
$=-A^{\mu \nu }B_{\mu \nu }$ (relabeling dummy indices)    (2)
$=-A^{\alpha \beta }B_{\alpha \beta }$ (relabeling dummy indices again)    (3)

Now, if you're not comfortable at this point with just relabeling indices, it's instructive to work out the expressions on both the L.H.S. and the R.H.S. of the equation:
$A^{\alpha \beta }B_{\alpha \beta } = -A^{\beta\alpha}B_{\beta\alpha}$    (1)

L.H.S.: $A^{\alpha \beta }B_{\alpha \beta } = A^{11}B_{11}+A^{12}B_{12}+A^{21}B_{21}+A^{22}B_{22}+\cdot \cdot \cdot$
R.H.S.: $-A^{\beta\alpha}B_{\beta\alpha} = -A^{11}B_{11}-A^{21}B_{21}-A^{12}B_{12}-A^{22}B_{22}+\cdot \cdot \cdot$
Rearranging terms and factoring out a negative one on the R.H.S. gives us:
$-A^{\beta\alpha}B_{\beta\alpha} = -1\ast \left [  A^{11}B_{11}+A^{12}B_{12}+A^{21}B_{21}+A^{22}B_{22}+\cdot \cdot \cdot \right ]$
Which shows that:
$-A^{\beta\alpha}B_{\beta\alpha}=-A^{\alpha \beta }B_{\alpha \beta }$

So, now we go back to (3) above:
$A^{\alpha \beta }B_{\alpha \beta }=-A^{\alpha \beta }B_{\alpha \beta }$
$\Rightarrow 2A^{\alpha \beta }B_{\alpha \beta }=0$
$\Rightarrow A^{\alpha \beta }B_{\alpha \beta } =0$    Q.E.D.

This Doesn't Taste Like Chicken

But, it is food for thought.

Do we come from the universe, or does the universe come from us?

Or just me?

Momentum-Space Wavefunction - Expectation Values: Quantum Mechanics (Griffiths)

In D. J. Griffiths' "Introduction to Quantum Mechanics: 1st Edition," Problem 3.51 on page 117 is a 3-star problem. In the second edition, it's problem 3.12 on page 109, and is not a starred problem at all. I attribute this to being given more information about $\Psi_{x,t}$ in terms of $\Phi_{p,t}$ in the second edition. In the first edition, we are working from knowledge of the form of $\Phi$ alone. Since one is a Fourier transform of the other and vice versa, we can convert the integral used to evaluate $< x>$  from one with an integrand function containing $\Psi$ and x to one containing $\Phi$ and the x operator.

$$< x> = \int \Phi^{*}\left ( -\frac{\hbar}{i}\frac{\partial }{\partial p} \right )\Phi dp$$

In the first edition, we're given

$$\Phi(p,t)=\frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}e^{-ipx/\hbar}\Psi(x,t)dx        (1)$$

We know $<x>=\int_{-\infty}^{\infty}x|\Psi(x,t)|^{2}dx$, so we just need to show that

$$\int \Phi^{*}\left ( -\frac{\hbar}{i}\frac{\partial }{\partial p} \right )\Phi dp=\int x|\Psi(x,t)|^{2}dx$$

Easier said than done, but here we go. It's not too bad, actually...

$$\int \Phi^{*}\left ( -\frac{\hbar}{i}\frac{\partial }{\partial p} \right )\Phi dp$$

Using (1) gives

$$ \int \left [ \frac{1}{\sqrt{2\pi \hbar}}\int e^{ipx^{'}/\hbar}\Psi^{*}dx^{'} \right ]\left ( -\frac{\hbar}{i} \frac{\partial }{\partial p}\right )\left [ \frac{1}{\sqrt{2\pi \hbar}}\int e^{-ipx/\hbar}\Psi dx \right ]dp$$

Simplifying a little...

$$\int \left [ \frac{1}{\sqrt{2\pi \hbar}}\int e^{ipx^{'}/\hbar}\Psi^{*}dx^{'} \right ]\left [ \frac{1}{\sqrt{2\pi \hbar}}\int i\hbar\frac{\partial }{\partial p} e^{-ipx/\hbar}\Psi dx \right ]dp        (2)$$

Consider the integrand function in the second set of brackets. In particular,  $i\hbar\frac{\partial }{\partial p} e^{-ipx/\hbar}$.

Using the fact that $\frac{\mathrm{d} }{\mathrm{d} p}\left ( e^{-ipx/\hbar} \right )= -\frac{ix}{\hbar} e^{-ipx/\hbar}$ we can rewrite (2) using $x e^{-ipx/\hbar}$ in place of $i\hbar \frac{\partial }{\partial p}\left ( e^{-ipx/\hbar} \right )$ ...

$$\int \left [ \frac{1}{\sqrt{2\pi \hbar}}\int e^{ipx^{'}/\hbar}\Psi^{*}dx^{'} \right ]\left [ \frac{1}{\sqrt{2\pi \hbar}}\int x e^{-ipx/\hbar} \Psi dx \right ]dp$$

Simplifying further and using Fubini's Theorem gives

$$\frac{1}{2\pi \hbar}\int \int \int \left [ e^{ipx^{'}/\hbar}\Psi^{*} x e^{-ipx/\hbar}\Psi \right ]dx^{'}dxdp     (3)$$

Letting $y = \frac{p}{\hbar}$ and simplifying the integrand further gives

$$\frac{1}{2\pi }\int \int \int \left [ e^{-i(x-x^{'})/y} \Psi^{*} x \Psi \right ]dx^{'}dxdy     (4)$$

Evaluating the y-integral first...

$$\frac{1}{2\pi }\int e^{-i(x-x^{'})/y}dy = \delta (x-x^{'})$$

If you're not comfortable with this, check it for yourself by using Plancherel's Theorem to determine the Fourier transform of $\delta(x)$.

Now we have

$$\int \int \delta(x-x^{'})  \Psi^{*}(x^{'},t) x \Psi (x,t) dx^{'}dx     (5)$$

But $ \int \delta(x-x^{'}) \Psi^{*}(x^{'},t) = \Psi^{*} (x,t)$  (see Eq. 3.129 in the first edition and Eq. 3.52 in the second edition), so we have

$$\int \Psi^{*}x \Psi dx = \int x |\Psi (x,t)|^{2}dx$$

QED

Proof of Plancherel's Theorem: Quantum Mechanics (Griffiths)

Problem 2.20 from chapter two of David J. Griffiths' "Introduction to Quantum Mechanics: 2nd Edition" takes the student through a step-by-step proof of Plancherel's Theorem:

$$f(x)=\frac{1}{\sqrt{2\pi }}\int_{-\infty}^{\infty}F(k)e^{ikx}dk \Leftrightarrow F(k)=\frac{1}{\sqrt{2\pi }}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$$

where F(k) is the Fourier transform of f(x), and f(x) is the inverse Fourier transform of F(k).

Step 1
First, Griffiths asks us to show that a function f(x) on the interval [-a, a] expanded as a Fourier series

$$f(x)=\sum_{n=0}^{\infty}[a_{n}sin(n\pi x/a)+b_{n}cos(n\pi x/a)]               [1]$$     

          

can be written equivalently as

$$f(x)=\sum_{n=-\infty}^{\infty}c_{n}e^{in\pi x/a}      [2]$$

and to write $c_{n}$ in terms of $a_{n}$ and $b_{n}$. 

If we look at n = 0, [1] --> $f(x) = b_{0}$

Now, use Euler's identity $e^{i\phi }=cos(\phi)+isin(\phi)$ to rewrite [1] as

$$f(x)=b_{0}+\sum_{n=1}^{\infty}\frac{a_{n}}{2i}(e^{in\pi x/a}-e^{-in\pi x/a})+\sum_{n=1}^{\infty}\frac{b_{n}}{2}(e^{in\pi x/a}+e^{-in\pi x/a})$$

And, rearranging this a little...

$$f(x)=b_{0}+\sum_{n=1}^{\infty}\left \{ \frac{a_{n}}{2i}+\frac{b_{n}}{2} \right \}e^{in\pi x/a}+\sum_{n=1}^{\infty}\left \{ \frac{-a_{n}}{2i} +\frac{b_{n}}{2}\right \}e^{-in\pi x/a}           [3]$$

Evaluating [3] with n = 1 gives

$$f(x)=\frac{1}{2}\left ( -ia_{1} +b_{1}\right )e^{i\pi x/a}+\frac{1}{2}\left ( ia_{1} +b_{1}\right )e^{-i\pi x/a}         [4]$$

Now, let's evaluate [2] with both n = -1 and n = 1

n = 1:   

$$f(x)=c_{1}e^{i\pi x/a}$$

n = -1: 

$$f(x)=c_{-1}e^{-i\pi x/a}$$

These two terms look like the two terms in [4] if $c_{1}$ = $\frac{1}{2}\left ( -ia_{1} +b_{1}\right )$ and  $c_{-1}$ = $\frac{1}{2}\left ( ia_{1} +b_{1}\right )$ (Remember, [3] sums on n from 1 to infinity, but [2] sums on n from negative infinity to positive infinity, so including the -n terms and the +n terms in [2] will give you the same two terms you get when evaluating [3] with only the +n terms.)

So, in general, $f(x)=\sum_{n=0}^{\infty}[a_{n}sin(n\pi x/a)+b_{n}cos(n\pi x/a)]$ can be written equivalently as

$$f(x)=\sum_{n=-\infty}^{\infty}c_{n}e^{in\pi x/a}$$
with $c_{n}$ = $\frac{1}{2}\left ( -ia_{n} +b_{n}\right )$, $c_{-n}$ = $\frac{1}{2}\left ( ia_{n} +b_{n}\right )$, and $c_{0}$ = $b_{0}$ 

QED

Step 2
Next, we are asked to show that

$$c_{n}=\frac{1}{2a}\int_{-a}^{+a}f(x)e^{-in\pi x/a}dx                [5]$$  

by using what Griffiths refers to as 'Fourier's Trick', which exploits the orthonormality of wavefunctions of different excited states.

The 'trick' is to multiply both sides of [2] by $\psi _{m}^{*}$ and integrate. Of course, in this case $\psi _{n}=e^{in\pi x/a}$.

$$f(x)=\sum_{n=-\infty}^{\infty}c_{n}e^{in\pi x/a}$$

$$\int_{-a}^{+a}\psi _{m}(x)^{*}f(x)dx = \sum_{n=-\infty}^{\infty}c_{n}\int_{-a}^{+a}\psi _{m}^{*}(x)\psi _{n}(x)dx=\sum_{n=-\infty}^{\infty}2ac_{n}\delta _{mn}=2ac_{m}$$

The orthonormality of $\psi_{m}^{*}$ and $\psi_{m}$ knocks out all the terms except for the n = m one, thus the appearance of the Kronecker delta.

So now we have

$$\int_{-a}^{+a}\psi_m^{*}(x)f(x)dx=2ac_{m}$$

Swapping indices (m for n) gives

$$\int_{-a}^{+a}\psi_n^{*}(x)f(x)dx=2ac_{n}$$

Finally, solving for $c_{n}$ and remembering that $\psi _{n}=e^{in\pi x/a}$ gives us

$$\frac{1}{2a}\int_{-a}^{+a}f(x)e^{-in\pi x/a}dx=c_{n}$$
QED

Step 3
Next, we are told to eliminate n and $c_{n}$ and use the new variables $k = (n\pi /a)$ and $F(k) = \sqrt{\frac{2}{\pi }}ac_{n}$ and show that [2] becomes

$$f(x)=\frac{1}{\sqrt{2\pi} }\sum_{n=-\infty}^{\infty}F(k)e^{ikx}\Delta k$$

and [5] (Step 2) becomes

$$F(k)=\frac{1}{\sqrt{2\pi} }\int_{-a}^{+a}f(x)e^{-ikx}dx$$

$F(k) = \sqrt{\frac{2}{\pi }}ac_{n}$ solved for $c_{n}$ gives $c_{n} = \frac{1}{a}F(k)\sqrt{\frac{\pi }{2}}$. Also, we know that $\Delta k=\frac{\Delta n\pi}{a}$ $\Delta n = 1$ so $\frac{\Delta k}{\pi }= \frac{1}{a}$

Substituting these results into [2] gives

$$f(x)=\frac{\Delta k}{\pi }\sqrt{\frac{\pi }{2}}\sum_{n=-\infty}^{\infty}F(k)e^{ikx}=\frac{1}{\sqrt{2\pi }}\sum_{n=-\infty}^{\infty}F(k)e^{ikx}\Delta k     QED$$

That takes care of f(x); Now for F(k)...

$$c_{n}=\frac{1}{2a}\int_{-a}^{+a}f(x)e^{-in\pi x/a}dx                [5]$$

But $c_{n} = \frac{1}{a}F(k)\sqrt{\frac{\pi }{2}}$, so [5] becomes

$$ \frac{1}{a}F(k)\sqrt{\frac{\pi }{2}}= \frac{1}{2a}\int_{-a}^{+a}f(x)e^{-in\pi x/a}dx$$

A little algebra and an appropriate substitution gives

$$ F(k) = \frac{1}{\sqrt{2\pi}}\int_{-a}^{+a}f(x)e^{-ikx}dx       QED$$


Final Step
To complete the proof of Plancherel's Theorem, we are asked to take the limit a $\rightarrow\infty$ for the two results from Step 3

1st, f(x)...

$$\lim_{\Delta k\rightarrow 0}f(x)=\lim_{\Delta k\rightarrow 0}\frac{1}{\sqrt{2\pi }}\sum_{n=-\infty}^{\infty}F(k)e^{ikx}\Delta k = \frac{1}{\sqrt{2\pi }}\int_{-\infty}^{\infty}F(k)e^{ikx}dk $$

Now, F(k)...

$$\lim_{a\rightarrow \infty}F(k) = \lim_{a\rightarrow \infty}\frac{1}{\sqrt{2\pi}}\int_{-a}^{+a}f(x)e^{-ikx}dx = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$$

So, finally we have Plancherel's Theorem as our result

$$f(x) = \frac{1}{\sqrt{2\pi }}\int_{-\infty}^{\infty}F(k)e^{ikx}dk \Leftrightarrow F(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$$

QED