Displacement operation acting on individual quadrature components

Hi all,

I have a naive understanding of how operators work and wondered if someone could help me. I have tried to understand this myself, but alas, I think my knowledge is too premature to understand what I am reading online. Is someone able to explain?

I want to perform the operation ##D(\alpha)^\dagger X D(\alpha)##, where ##X## is one of the quadrature components ##X = \frac{1}{\sqrt{2}} (a + a^\dagger)##. I know the following:

$$
D(\alpha)^\dagger a D(\alpha) = a + \alpha
$$

Does this mean ##D(\alpha)^\dagger a^\dagger D(\alpha) = a^\dagger + \alpha^*##?

If so, then:

$$
D(\alpha)^\dagger X D(\alpha) = \frac{1}{\sqrt{2}} D(\alpha)^\dagger (a + a^\dagger) D(\alpha) = \frac{1}{\sqrt{2}} (a + a^\dagger + 2|\alpha|)
$$

This is not correct. I would have expected something like ##\frac{1}{\sqrt{2}} (a + a^\dagger + |\alpha|)##.

I am struggling even more to work out ##P##, where ##P = \frac{i}{\sqrt{2}} (a - a^\dagger)##. I get:

$$
D(\alpha)^\dagger P D(\alpha) = \frac{1}{\sqrt{2}} D(\alpha)^\dagger (a - a^\dagger) D(\alpha) = \frac{1}{\sqrt{2}} (a - a^\dagger)
$$

and it remains the same, which is not correct. I would also expect ##\frac{1}{\sqrt{2}} (a - a^\dagger + |\alpha|)##.

Can someone tell me where I am going wrong?

I appreciate any help you can provide.

Since operator a is not Hermitian, I am not sure and accoustomed to translational operation on it. Is it an exercise on textbook ?

waadles said:

Hi all,

I have a naive understanding of how operators work and wondered if someone could help me. I have tried to understand this myself, but alas, I think my knowledge is too premature to understand what I am reading online. Is someone able to explain?

I want to perform the operation ##D(\alpha)^\dagger X D(\alpha)##, where ##X## is one of the quadrature components ##X = \frac{1}{\sqrt{2}} (a + a^\dagger)##. I know the following:

$$
D(\alpha)^\dagger a D(\alpha) = a + \alpha
$$

Does this mean ##D(\alpha)^\dagger a^\dagger D(\alpha) = a^\dagger + \alpha^*##?

That follows by taking the Hermitian conjugate of the equation.

waadles said:

If so, then:

$$
D(\alpha)^\dagger X D(\alpha) = \frac{1}{\sqrt{2}} D(\alpha)^\dagger (a + a^\dagger) D(\alpha) = \frac{1}{\sqrt{2}} (a + a^\dagger + 2|\alpha|)
$$

Note that ##\alpha + \alpha^* = 2Re(\alpha)##.

waadles said:

This is not correct. I would have expected something like ##\frac{1}{\sqrt{2}} (a + a^\dagger + |\alpha|)##.

Why did you expect that?

waadles said:

I am struggling even more to work out ##P##, where ##P = \frac{i}{\sqrt{2}} (a - a^\dagger)##. I get:

$$
D(\alpha)^\dagger P D(\alpha) = \frac{1}{\sqrt{2}} D(\alpha)^\dagger (a - a^\dagger) D(\alpha) = \frac{1}{\sqrt{2}} (a - a^\dagger)
$$

In this case you should use ##\alpha - \alpha^* = 2iIm(\alpha)##.

PeroK said:

In this case you should use α−α∗=2iIm(α).

May we say about meaning of translation of a as
[tex]\alpha=\frac{d_x}{\sqrt{2}}-i\frac{d_p}{\sqrt{2}}[/tex]
where d_x and d_p are displacement in coordinate and momentum space ?
If so it holds for the translated Hamiltoian, i.e.
[tex](a^{\dagger} +\alpha^*)(a+\alpha)+\frac{1}{2}[/tex]? It seems not obvious to me.