So my lecture posed and integral of the form:
$$\langle x\rangle=\int_{-\infty}^\infty \psi^*(y)\delta(y-x)\psi(y) \, dy = \psi^(x) \psi(x)$$
and then looked at the equation and said have I gone wrong here, so before he could figure out if he was incorrect the lecture was over and posed it as a problem for the class.
Now I am not sure if he is wrong/right but I am under the assumption at the moment that this is correct and if but what I can understand is how $\psi(x)\psi^*(x)$ can transform to $\psi(y)\psi^*(y)$
My working are as follows:
As the dirac delta $$\int _{-\infty}^\infty \delta(x-y) \, dx=1 \tag{1}$$
then
$$\int_{-\infty}^\infty \delta(y-x) \, dy=1\tag{[2]}$$
then $(1)=(2).$
$$\int _{-\infty }^\infty \delta(x-y) \, dx=\int_{-\infty}^\infty \delta(y-x) \, dy$$
but after this I just cant see how the functions match each other and to be honest I can see why the Dirac delta function has been used, I have tried looking for question to get an idea on what the is applied to but so far I have come up trumps. If possible could someone please explain, how and why the relationship comes about.
Best Answer
The dirac delta "function" works as an identity operation so that $$f(x)=\int_a^b\delta(x-y)f(y)\,dy$$ so in some kind of way it can be compared to the identity matrix.
In your case $f(x)=\psi(x)\psi^*(x)$ and $f(y)=\psi(y)\psi^*(y)$.
What may be helpful to you is that the dirac delta "function" has very interesting properties one of which is symmetry so that $\delta(x)=\delta(-x)$ or $$\delta(y-x)=\delta(-[y-x])=\delta(x-y)$$
Thats why you would get $$\int_{-\infty}^\infty f(x)\delta(x-y) \, dx=\int_{-\infty}^\infty f(x)\delta(y-x) \, dx $$
Then you can just change the name of the variables.