Normalized wave functions in position and momentum space

Question

Using the following expression for the Dirac delta function: $$\delta(k-k')=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(k-k')x}\mathrm{d}x$$

show that if $\Psi(x,t)$ is normalized at time $t=0$, then the corresponding momentum space wave function $\Phi(p_x,t)$ is also normalized at time $t=0$.

Since $\Psi(x,0)$ is normalized, we know $$\int_{-\infty}^{\infty}\Psi^*(x,0)\Psi(x,0)\mathrm{d}x=1$$

By definition of the momentum space wave function, $$\Phi(p_x,0)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}e^{\frac{i}{\hbar}p_xx}\Psi(x,0)\mathrm{d}x$$ and $$\Phi^*(p_x,0)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}e^{\frac{i}{\hbar}p_xx}\Psi^*(x,0)\mathrm{d}x$$

To check if $\Phi(p_x,0)$ is normalized, we'd like to check if $\Phi(p_x,0)\Phi^*(x,0)$ integrates to $1$ over all values of $p_x$, i.e. we need to evaluate $$\Phi^*(p_x,0)\Phi(p_x,0)=\frac{1}{2\pi\hbar}\left(\int_{-\infty}^{\infty}e^{\frac{i}{\hbar}p_xx}\Psi^*(x,0)\mathrm{d}x\right)\left(\int_{-\infty}^{\infty}e^{\frac{i}{\hbar}p_xx}\Psi(x,0)\mathrm{d}x\right)$$

I remembered the method for evaluating the Gaussian integral, where a product of integrals is expressed as a double integral, so here's my attempt at that: $$\Phi^*(p_x,0)\Phi(p_x,0)=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{\frac{i}{\hbar}p_xy}\Psi^*(y,0)e^{\frac{i}{\hbar}p_xx}\Psi(x,0)\mathrm{d}y\mathrm{d}x$$

$$=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{2i\frac{p_x}{\hbar}yx}\Psi^*(y,0)\Psi(x,0)\mathrm{d}y\mathrm{d}x$$

I have no idea if this is the correct path, and while I can see similarities between the delta function definition that I've been given and what I have now, I'm not exactly sure how to make the leap and start making use of some properties of the Dirac delta.

Any hints are appreciated!

Answer 1

Using the following expression for the Dirac delta function: $$\delta(k-k')=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(k-k')x}\mathrm{d}x$$

show that if $\Psi(x,t)$ is normalized at time $t=0$, then the corresponding momentum space wave function $\Phi(p_x,t)$ is also normalized at time $t=0$.

Good.

By definition of the momentum space wave function, $$\Phi(p_x,0)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}e^{\frac{i}{\hbar}p_xx}\Psi(x,0)\mathrm{d}x$$ and $$\Phi^*(p_x,0)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}e^{\frac{i}{\hbar}p_xx}\Psi^*(x,0)\mathrm{d}x$$

No. $$\Phi^*(p_x,0)\neq\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}e^{\frac{i}{\hbar}p_xx}\Psi^*(x,0)\mathrm{d}x$$

Just write out the real and imaginary parts of the previous integral and flip the sign of the imaginary one. And the $x,$ $dx$ and $y,$ $dy$ are dummy variables, they are just the name of something that changes and so can be called anything. Call the second one $dy$ instead of $dx$ to make your life easier later. But keep the $p_x$ as $p_x$ since that isn't a dummy. It isn't changing, it's the fixed value on the left hand side over in $\Phi^*(p_x,0).$

To check if $\Phi(p_x,0)$ is normalized, we'd like to check if $\Phi(p_x,0)\Phi^*(x,0)$ integrates to $1$ over all values of $p_x$,

Another mistake, this one might just be a typo.

$$=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{2i\frac{p_x}{\hbar}yx}\Psi^*(y,0)\Psi(x,0)\mathrm{d}y\mathrm{d}x$$

Also wrong, and not just because of previous mistakes carrying forward but because of a new mistake. $e^Ae^B=e^{A+B}$ and you can check that $A,$ $B,$ and $A+B$ all have the same units, dimensionless units.

I'm not exactly sure how to make the leap and start making use of some properties of the Dirac delta.

After you get it to look exactly like the delta you can use some change of variables to use the one property of the delta.

Answer 2

To calculate $\langle\psi | \psi\rangle$ we have, using the expansion of the identity onto the momentum basis: $$ \langle\psi | \psi\rangle = \int\textrm{d}p\,\langle\psi|p\rangle\langle p|\psi\rangle = \int\textrm{d}p |\psi(p)|^2 $$ to be proven to be 1. Let us insert the identity as expanded onto $x$ and $x'$ $$ \int\textrm{d}p\,\textrm{d}x\,\textrm{d}x'\,\langle\psi|x\rangle\langle x|p\rangle\langle p|x'\rangle\langle x'|\psi\rangle. $$ Since $\langle x|p\rangle = \textrm{e}^{-ipx}$ we have $$ \int \textrm{d}x\,\textrm{d}x'\,\langle\psi|x\rangle\langle x'|\psi\rangle\int\textrm{d}p\,\textrm{e}^{-i(x-x')p} $$ integration in $\textrm{d}p$ gives back $\delta(x-x')$ and thus $$ \int \textrm{d}x\,\langle\psi|x\rangle\int\textrm{d}x'\langle x'|\psi\rangle\,\delta(x-x')=\int \textrm{d}x\,\langle\psi|x\rangle\langle x|\psi\rangle = 1. $$ If the wave function is normalised onto the momentum basis, it is also normalised onto the position basis and viceversa due to the fact that $\langle x|p\rangle$ is essentially a plane wave that integrates to the $\delta$ (and this follows from the canonical commutation relations). I have assumed that one can by all means switch integrals inside and outside; furthermore, all kets are supposed to hold so at any given time $t$.

P. S. I might have forgotten all the $2\pi$ here and there or, as my old professor of quantum mechanics used to say, we have set $2\pi = 1$.