Question

Can the electric dipole moment and the energy of a molecule be specified simultaneously? (QM)

Answer 1

I just consider an example with the electric dipole moment of a charge distribution of a proton at the position origin and an electron at the position I call \(\bf r\). The electric dipole moment is then defined as:
\[\large \bf \mu \it=Q ~ \bf r\]
Where Q in this case is the negative fundamental charge \(e\) and we consider this example 1 dimensional, so for my example:
\[\large \bf \mu \it=-e ~ x\]
I'm gonna try use the general uncertainty principle to check for simultaneous observation, so I'm gonna check if they electric dipole moment operator and hamiltonian operator commute:
\[\large \hat{\bf\mu}=-e ~ x \times\]
\[\large \hat{H}=\hat{E _{k}}+\hat{V}\]
I'm gonna spit up the hamiltonian and start by checking the potential energy:
\[\large \hat{\mu}~\hat{V} \Psi=-e ~ x \times V \times \Psi\]
\[\large \hat{V}~ \hat{\mu} \Psi=V \times (-e) ~ x \times \Psi\]
We see the potential energy operator and electric dipole moment commute, as the commutator equals 0:
\[\large \left[ \hat{\mu},\hat{V} \right]=\hat{\mu}~\hat{V} - \hat{V}~ \hat{\mu} =0\]
So the problem most be the kinetic energy operator \(\hat{E_{k}}\) and electric dipole moment. Defining the kinetic energy operator due to postulate III:
\[\large E _{k}= \frac{ p^2 }{ 2m _{e} } \to \hat{E _{k}}=\frac{ \hat{p}^{2} }{ 2m _{e} }=-\frac{ \hbar ^{2} }{ 2m _{e} } \frac{ d ^{2} }{ dx ^{2} }\]
Checking for if the operators commute:
\[\large \hat{\mu} \hat{E _{k}} \Psi=-ex \times (-\frac{ \hbar ^{2} }{ 2m _{e} }\frac{ d ^{2} }{ dx ^{2} }\Psi)\]
\[\large \hat{E _{k}} \hat{\mu} \Psi=-\frac{ \hbar ^{2} }{ 2m _{e} } \frac{ d ^{2} }{ dx ^{2} }(-ex \Psi)\]
I get the commutator to the following (no sure it is 100% correct):
\[\large \left[ \hat{\mu}, \hat{E _{k}} \right] \Psi=-\frac{ \hbar ^{2}e }{ 2m _{e} }\left( x \frac{ d ^{2} }{ dx ^{2} }- \frac{ d ^{2} }{ dx ^{2} }x \right) \Psi\]

Answer 2

Or in fact I am sure it is correct (but remove the minus sign please), it is the following I'm unsure about, next part is just math:
\[\large \frac{ d ^{2} }{ dx }(x \Psi)=\frac{ d }{ dx }\left( \frac{ d }{ dx } x \Psi \right)=\frac{ d }{ dx }\left( \Psi+x \frac{ d \Psi }{ dx } \right)\]
\[\large \frac{ d }{ dx }\left( \Psi +x \frac{ d \Psi }{ dx }\right)=\frac{ d \Psi }{ dx } + \frac{ d }{ dx }\left( x \frac{ d \Psi }{ dx } \right)=\frac{ d \Psi }{ dx }+\frac{ d \Psi }{ dx }+x \frac{ d ^{2} \Psi }{ dx }\]
\[\large \frac{ d ^{2} \Psi }{ dx }\left( x \Psi \right)=2 \frac{ d \Psi }{ dx }+ x \frac{ d ^{2} \Psi }{ dx }\]
Substitute into the last expression in last reply (with one with the minus mistake):
\[\large \left[ \hat{\mu}, \hat{E _{k}} \right] \Psi= \frac{ \hbar ^{2} e}{ 2m _{e} }\left( x \frac{ d ^{2 }\Psi }{ dx }-2\frac{ d \Psi }{ dx }+x \frac{ d ^{2} \Psi }{ dx } \right)=-\frac{ \hbar ^{2}e }{ m _{e} }\frac{ d \Psi }{ dx }\]
The expression you can rewrite so it consist of the linear momentum operator:
\[\large \left[ \hat{\mu}, \hat{E _{k}} \right] \Psi=-\frac{ i \hbar ^{2}e }{ i m _{e} }\frac{ d }{ dx } \psi=\frac{ -i \hbar e }{ m _{e} } \hat{p _{x}} \Psi\]

Answer 3

So to return to what matters, the commutator for the electric dipole moment and the hamiltonian operator:
\[\large \left[ \hat{\mu}, \hat{H} \right]=\left[ \hat{\mu}, \hat{K _{e}}+\hat{V} \right]=\left[ \hat{\mu}, \hat{K _{e}} \right]+\left[ \hat{\mu}, \hat{V} \right]=\frac{ -i \hbar e }{ m _{e} } \hat{p _{x}}+0=\frac{ -i \hbar e }{ m _{e} } \hat{p _{x}}\]
So the commutator does not equal 0 and therefore the electric dipole moment and energy most be complementary observables.
If using the general uncertainty principle I get the following:
\[\large \Delta \mu ~ \Delta E \ge \frac{ 1 }{ 2 } \left| \langle \left[ \hat{\mu},\hat{H} \right]\rangle \right| \ge \frac{ 1 }{ 2 } \left| \langle \frac{ -i \hbar e }{ m _{e} } \hat{p _{x}}\rangle \right| \ge \frac{ \hbar e }{ 2m _{e} } \langle \hat{p _{x}} \rangle\]
But does this mean that so long linear momentum operator equal 0, then you can determine the electric dipole moment and the energy even though the operators are complementary?

Answer 4

And something else I'm wondering about:
If you choose to use postulate III you should be able to write the electric dipole operator as:
\[\large \hat{\mu}=-e ~ \hat{x}\]
We then get the following uncertainty principle:
\[\Delta \mu ~ \Delta E = -e \Delta x \Delta E \ge - \frac{ 1 }{ 2 }e \left| \langle \left[ \hat{x},\hat{H} \right]\rangle \right| \ge -\frac{ 1 }{ 2 }e \left| \langle \frac{ i \hbar }{ m _{e} } \hat{p _{x}}\rangle \right| \ge- \frac{ \hbar e }{ 2m _{e} } \langle \hat{p _{x}} \rangle\]
When comparing the two uncertainty principles I get to the conclusion:
\[\large \langle \hat{p _{x}} \rangle = - \langle \hat{p _{x}} \rangle\]

Answer 5

So yeah, basically you the kinetic energy and position operators don't commute, but potential energy and position do commute. No surprise there that since the Hamiltonian is just a linear combination of the two that total energy doesn't commute with position (AKA a electric dipole moment operator with a constant divided out basically).

The significance of <P>=-<P> is simply that you know the speed but not direction along the x-axis.

Answer 6

Also, you can never have a particle with 0 velocity, it's part of why it seems like you can find break the uncertainty principle and find the electric dipole even though they are complementary.

Answer 7

Well what if we assume they do have velocity, but just no net linear momentum?

Answer 8

Also wrong of me when I said that momentum = 0. i should say \( \langle \hat{p_{x}} \rangle =0 \)

Answer 9

@Kainui but thank you for clearing out the \(\langle\hat{p_{x}}\rangle=-\langle\hat{p_{x}}\rangle\). Did not think about that we could explain it using superpositions.

Answer 10

Not sure how you can have velocity but no net linear momentum. What would that even look like?

Answer 11

We could make them collide?