Calculating ##\langle zHz \rangle## in different ways

[hmparticle9]

Homework Statement

Calculating $\langle zHz \rangle$ in the ground state of hydrogen.

Relevant Equations

$H = \frac{1}{2m}(p_x^2 + p_y^2 + p_z^2) + V$

We can calculate $\langle zHz \rangle$ by noting that $[z,[H,z]] = 2zHz - z^2H - Hz^2$. The author says this is the quick way: it takes "four lines". Embarrassingly, it took me way longer than that. For instance, in the calculation of $[H,z]$, he says that
$$[p_x^2 + p_y^2 + p_z^2, z] = [p_z^2, z].$$
I had to prove that to myself using test functions, which took about four lines by itself. There is no commentary in the solution to this problem but am I correct in saying that since $p_x, p_y$ 'get rid of' $z$ we can proceed as the author did? The advantage of doing it my way is that you immediately get
$$\frac{1}{2m}[p_z^2, z] = -\frac{ih}{m}p_z$$

Regardless, I solved the problem in the end.

Now onto the main part of my question. In the statement of the problem he mentions that $\langle zHz \rangle$ can be calculated the "long way". I am not sure what he means by this, but I would like to find out. Maybe we will start like this (an operator acts on everything to the right of it):

$$ \langle zHz \rangle = \int \psi^* zH(z \psi r^2 \sin \theta) \text{ d}r \text{ d}\theta \text{ d} \psi$$

I don't think this is correct as we are going to introduce derivatives of $\theta$ w.r.t $x$ etc.

 

[ergospherical]

That's the correct "long way" (or at least, one of them). How you go about evaluating that is up to you... but there's certainly a few different routes you could try. Since this post is in the Homework section, you should make an attempt before we can help further. Here are a couple of suggestions:

(i) Since you're dealing with the ground state of Hydrogen, you might happen to already know the 1$s$ wave function: $\psi(r) = \tfrac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0}$, along with the potential $V(r) = -\tfrac{\hbar^2}{m_e a_0 r}$. You can directly evaluate your expression.

(ii) You can also do it without knowing the explicit ground state wave function, but you'll need to use the Schrödinger equation. In particular, you could compare the expression you get by multiplying $H\psi = E \psi$ (for your $H = -\tfrac{\hbar^2}{2m} \nabla^2 + V$) by $z^2 \psi^{*}$ and integrating, to the integral you wrote down.

 

[JimWhoKnew]

I am not sure what he means by this, but I would like to find out. Maybe we will start like this (an operator acts on everything to the right of it):

$$ \langle zHz \rangle = \int \psi^* zH(z \psi r^2 \sin \theta) \text{ d}r \text{ d}\theta \text{ d} \psi$$

The Jacobian determinant $r^2\sin\theta$ should be outside the brackets. That is:$$ \langle zHz \rangle = \int \psi^* (zH(z \psi)) r^2 \sin \theta \text{ d}r \text{ d}\theta \text{ d} \psi$$I also recommend using capital letters for operators, to distinguish them from variables.

 

[kuruman]

I also recommend that you use $d\phi$ instead of $d\psi$ where appropriate to avoid confusion with the wavefunction.

 

[hmparticle9]

I really doubted myself because I found the "long way" to actually be shorter than the "short way". Thanks again guys.

 

[ergospherical]

Right. The "quick way" often means "here's this esoteric trick that my professor taught me (and his professor taught him, etc.) that you'll probably never come up with unless you mess around with algebra for three weeks...."

 

[PeroK]

Right. The "quick way" often means "here's this esoteric trick that my professor taught me (and his professor taught him, etc.) that you'll probably never come up with unless you mess around with algebra for three weeks...."

I thought it was a reference to Arsenic and Old Lace, where Peter Lorre's character implores his accomplice to do the killing "ze kveek vay", because the slow way is so messy!

 

[kuruman]

I thought it was a reference to Arsenic and Old Lace, where Peter Lorre's character implores his accomplice to do the killing "ze kveek vay", because the slow way is so messy!

Actually it's a double reference. Doing it "zee kveek vay" appears in Griffiths' QM book who attributes it to Peter Lorre in Arsenic and Old Lace. Great film!

 

[JimWhoKnew]

I really doubted myself because I found the "long way" to actually be shorter than the "short way". Thanks again guys.

Since $[z,[H,z]]$ is a constant (times the identity operator), and $\langle \psi_{100}|H=E_0\langle \psi_{100}|$ , the only integration you need in the "short way" is for $\langle z^2 \rangle$ . It should be slightly easier than the "long way", even if you zig-zag between the Cartesian expressions for $H$ & $Z$ and the spherical-polar integration.