Why Does My Calculation Have an Extra Factor of 2?

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RealKiller69
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Homework Statement
Calculate. <2,0,0|&(x)P_z+P_z&(x)|2,1,0>
Where |n,l,m> are the hydrogen atom eigenstates, &(x) is dirac delta and x is the position operator . It should be -iħ/(16pia_0^4) a_0=ħ^2/me^2 .
Relevant Equations
Not many
15745216454334312777540192084427.jpg

My solution is not the same, it has an additional *2 factor and i can't see any mistakes.
 
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I will chime into say yes, please use the LaTeX guide to format your question in readable format. That doesn't describe every LaTeX feature, but most of what you need to ask your question. But also to give some examples. It's not that hard to learn.

Here is your question with the LaTeX.

Calculate. <2,0,0|&(x)P_z+P_z&(x)|2,1,0>

Use \delta for a lower case ##\delta## symbol. The rest of that expression will work as is. Embed it between ##\text{##}## symbols to invoke the LaTeX interpreter. Thus, you would type this:
##\text{##}##<2,0,0|\delta(x)P_z+P_z \delta(x)|2,1,0>##\text{##}##
to get this:
##<2,0,0|\delta(x)P_z+P_z \delta(x)|2,1,0>##

It should be -iħ/(16pia_0^4) a_0=ħ^2/me^2

Is that all meant to be one equation? Perhaps this one?
##\left [ -i \hbar / (16 \pi a_0^4) \right ] a_0 = \hbar^2 / me^2##
Again, what you wrote is almost correct LaTeX syntax. The LaTeX looks like this:
##\#\### \left [ -i \hbar / (16 \pi a_0^4) \right ] a_0 = \hbar^2 / me^2##\#\###

Prefixing brackets with "\left" and "\right" creates stretchy brackets that stretch vertically enough to match whatever is inside.