Could an operator act on a bra vector?

[Haorong Wu]

Homework Statement

Show that
$a^{\dagger} \left | \alpha \right > \left < \alpha \right | = \left ( \alpha ^{*} + \frac {\partial} {\partial \alpha} \right ) \left | \alpha \right > \left < \alpha \right |$

Relevant Equations

$\left | \alpha \right >$ is a coherent state, and $a\left | \alpha \right >=\alpha \left | \alpha \right >$.
Also $\left | \alpha \right > = e^{- \left | \alpha \right | ^2 /2} \sum_{n=0}^{\infty} \frac {\alpha ^n} {\sqrt{n!}} \left | n \right >$

I am confused about the problem. I thought operators do not act on bra vectors, and the problem is equivalent to
$a^{\dagger} \left | \alpha \right > = \left ( \alpha ^{*} + \frac {\partial} {\partial \alpha} \right ) \left | \alpha \right >$. Then, strangely, $\left < \alpha \right |$ is redundant. So I think, $\left < \alpha \right |$ is included because the operator $a^\dagger$ has some effects on it.

On the other hand, I could not solve the problem. I tried to insert the completeness ralation so that
$a^{\dagger} \left | \alpha \right > \left < \alpha \right | =\frac 1 {\pi} \int d^2 \alpha ^{'} a^{\dagger} \left | \alpha^{'} \right > \left < \alpha^{'} \right |\left | \alpha \right > \left < \alpha \right |$,
and tried to expand $\left | \alpha \right > = e^{- \left | \alpha \right | ^2 /2} \sum_{n=0}^{\infty} \frac {\alpha ^n} {\sqrt{n!}} \left | n \right >$.

But neither method can give any terms including $\frac {\partial} {\partial \alpha}$

Meanwhile, I am not sure how to do partial differential with $\alpha$ which is a complex number.

By the way, what is the meaning of $\Box ^ 2$ in $\Box ^2 E = - \mu _0 P$?

Thanks!

 

[anuttarasammyak]

and the problem is equivalent to
a†|α⟩=(α∗+∂∂α)|α⟩.

I think so. Did you try calculating
$$\frac{\partial}{\partial \alpha} [e^{-\alpha\alpha^*/2}\alpha^n]\frac{|n>}{\sqrt{n!}}$$?

 

[DrClaude]

I thought operators do not act on bra vectors,

Well, operators can act on bra vectors from the right, but I don't think this is what you meant.

and the problem is equivalent to
$a^{\dagger} \left | \alpha \right > = \left ( \alpha ^{*} + \frac {\partial} {\partial \alpha} \right ) \left | \alpha \right >$. Then, strangely, $\left < \alpha \right |$ is redundant. So I think, $\left < \alpha \right |$ is included because the operator $a^\dagger$ has some effects on it.

It is equivalent. The bra is "passive" here. My guess is that this relation will be useful later.

I tried to insert the completeness ralation so that
$a^{\dagger} \left | \alpha \right > \left < \alpha \right | =\frac 1 {\pi} \int d^2 \alpha ^{'} a^{\dagger} \left | \alpha^{'} \right > \left < \alpha^{'} \right |\left | \alpha \right > \left < \alpha \right |$,

This is not useful here. The point of using the completeness relation is when you have an expression of the type $\hat{A} | \psi \rangle$ which you might be able to solve by replacing the $| \psi \rangle$ with eigenkets of $\hat{A}$, for which you know how the operator acts on them.

and tried to expand $\left | \alpha \right > = e^{- \left | \alpha \right | ^2 /2} \sum_{n=0}^{\infty} \frac {\alpha ^n} {\sqrt{n!}} \left | n \right >$.

This is the right approach. Of course, the $\frac {\partial} {\partial \alpha}$ won't appear by itself, but it is useful to consider how that derivative applies to $e^{c \alpha}$.

But neither method can give any terms including $\frac {\partial} {\partial \alpha}$

Meanwhile, I am not sure how to do partial differential with $\alpha$ which is a complex number.

What is $\frac {\partial} {\partial x}$ of $f(x) = x^n$?

By the way, what is the meaning of $\Box ^ 2$ in $\Box ^2 E = - \mu _0 P$?

It is the d'Alembertian.

 

[Haorong Wu]

I think so. Did you try calculating
$$\frac{\partial}{\partial \alpha} [e^{-\alpha\alpha^*/2}\alpha^n]\frac{|n>}{\sqrt{n!}}$$?

Hi, @anuttarasammyak . I have tried it. I was stuck in calculating the partial derivative $\frac {\partial \alpha ^*} {\partial \alpha}$.
Because $\alpha ^ *$ is not analytic, then the derivative $\frac {d \alpha ^*} {d \alpha}$ does not exist. I do not know how to deal with this situation.

 

[Haorong Wu]

Many thanks, @DrClaude .

I guess that, operator $a^\dagger$ may not act on the bra, but the partial do act on it. I will try it hard.

Also, since $\alpha ^ *$ is not analytic, its derivative does not exist, so I do not know how to proceed next. Should I ignore it?

 

[anuttarasammyak]

I was stuck in calculating the partial derivative ∂α∗∂α.

Thanks. So the answer suggests us
$$\frac{\partial}{\partial \alpha}|\alpha><\alpha|=[\frac{\partial}{\partial \alpha}|\alpha>]<\alpha|\ +\ |\alpha>[\frac{\partial}{\partial \alpha}<\alpha|]$$, partial derivative works not on ket only but on whole operator, and
$$\frac{\partial}{\partial \alpha}(\alpha \alpha^*)=\alpha^*$$
,alpha* is independent parameter as for partial derivative of alpha.

 

[George Jones]

I guess that, operator $a^\dagger$ may not act on the bra, but the partial do act on it.

Yes, I think so.

I do not have time to look at this today (I am in the middle of a bunch of work work), but here is a suggestion (which may or may not be helpful). You are trying to prove an equation involving operators, and one way to do this is to show that when both sides operate on an arbitrary ket, they give the same result. Since they form basis, it is sufficient to operate on an arbitrary energy eigenket $\left| m \right>$. This should get rid of one of the sums.

Again, I have no idea if this is useful.

 

[George Jones]

After going back to my real work, this was still in the back of my mind. Maybe take this idea one step further and calculate matrix elements (with respect to an energy eigenbasis) of both operators by sandwiching the operators on both sides of the desired equation between the bra $\left< n \right|$ and the ket $\left| m \right>$. Let $a^\dagger$ operate on the bra $\left< n \right|$.

 

[anuttarasammyak]

Because α∗ is not analytic, then the derivative dα∗dα does not exist.

What is ∂∂x of f(x)=xn?

Say $f(z)=z^*$, making conjugate complex,
f(z) does not satisfies Cauchy-Riemann relation, i.e.
$$z=x+iy,f(z)=x-iy$$
$$\frac{\partial x}{\partial x} \neq \frac{\partial (-y)}{\partial y}$$
So it does not differentiable.

Say
$$f'(z)=\lim_{\zeta \rightarrow 0}\frac{z^*+\zeta^*-z^*}{\zeta}=\lim_{\zeta \rightarrow 0}e^{-i \ arg(\zeta)}$$
It does not depend on value of z and is a phase factor of any chosen infinitesimal approach. If we can take "averaged" phase factor
$$<e^{-i \ arg(\zeta)}>=0$$ then
$$f'(z)=\frac{dz^*}{dz}=0$$
Such an extended meaning of partial differentiation makes sense?

I should appreciate it if someone who knows mathematics better correct/comment on it.
　

 

[George Jones]

Homework Statement:: Show that
$a^{\dagger} \left | \alpha \right > \left < \alpha \right | = \left ( \alpha ^{*} + \frac {\partial} {\partial \alpha} \right ) \left | \alpha \right > \left < \alpha \right |$

After going back to my real work, this was still in the back of my mind. Maybe take this idea one step further and calculate matrix elements (with respect to an energy eigenbasis) of both operators by sandwiching the operators on both sides of the desired equation between the bra $\left< n \right|$ and the ket $\left| m \right>$. Let $a^\dagger$ operate on the bra $\left< n \right|$.

Now I've had a chance to calculate! This works.