Antilinear Operators

[hokhani]

TL;DR

The correctness of an equation for antilinear operators

An antilinear operator $\hat{A}$ can be considered as, $\hat{A}=\hat{L}\hat{K}$, where $\hat{L}$ is a linear operator and $\hat{K} c=c^*$ ($c$ is a complex number). In the Eq. (26) of the text https://bohr.physics.berkeley.edu/classes/221/notes/timerev.pdf the equality $(\langle \phi |\hat{A})|\psi \rangle=[ \langle \phi|(\hat{A}|\psi \rangle)]^*$ is given but I think this equation is not correct within a minus sign. For example, in the Hilbert space of spin up and down, having $\hat{L}=\hat{\sigma_y}$ and $|\psi\rangle=\psi_1 |+\rangle +\psi_2 |-\rangle$ and $|\phi\rangle=\phi_1 |+\rangle +\phi_2 |-\rangle$ we have: $\langle \phi | (\hat{A} |\psi\rangle)=-i\phi_1^* \psi_2^*+i\phi_2^*\psi_1^*$ and $(\langle \phi|\hat{A})|\psi \rangle=i\phi_2 \psi_1 -i\phi_1 \psi_2$ which gives $(\langle \phi |\hat{A})|\psi \rangle=-[ \langle \phi|(\hat{A}|\psi \rangle)]^*$. I appreciate any help.

 

[QuarkyMeson]

I'm not seeing the minus sign, basically in your example you need to apply the given definition:
$$\hat{A} = \hat{L} \hat{K}$$
$$\hat{L} = \sigma_y = (0, -i; i, 0)$$
$$ |\phi> = \phi_1|+> + \phi_2|->$$
$$ |\psi> = \psi_1|+> + \psi_2 |->$$
$$<\phi|(\hat{A}|\psi>) = -i\phi_1^*\psi_2^*+i\phi_2^*\psi_1^*$$
Conjugated:
$$ [<\phi|(\hat{A}|\psi>)]^* = i\phi_1\psi_2-i\phi_2\psi_1$$

So I think you're fine to here, your problem is you need to apply the definition to figure out what $(<\phi|\hat{A})$ is.

$$<\phi|\hat{A} = -i\phi_2<+| + i\phi_1<-|$$

 

[div_grad]

@hokhani I wouldn't focus too much on Eq.(26), as it seems that it's just a partiular definition the author chose to adpot in order to derive the main result, Eq.(32). In this sense, you may very well just adopt Eq.(32) as the definition of an anti-unitary operator.

However, maybe it will be more satisfactory for you to derive Eq.(32) using the definition of the form $A=LK$, but for an anti-unitary (and not merely anti-linear) operator. So let $A=UK$ be an anti-unitary operator, which is construced from a unitary operator $U$ and the "complex-conjugation'' operator $K$. Let $\phi$ and $\psi$ be state vectors in the complex Hilbert space $H$, and let $\langle\cdot,\cdot\rangle: H\times H \rightarrow \mathbb{C}$ be the inner product defined as (for simplicity we choose this; you can also define inner products in other ways, depending on the underlying representation of the Hilbert space)
$$
\langle \phi, \psi \rangle = \sum_{i}\phi_i^* \psi_i \rm{,}
$$
where $\phi_i, \psi_i \in \mathbb{C}$ are the components of the Hilbert-space state vectors $\phi$ and $\psi$ (note that the components are complex numbers). The unitary operator $U$, by definition, satisfies
$$
\langle U\phi, U\psi \rangle = \langle \phi, U^\dagger U \psi \rangle = \langle \phi, \psi \rangle \rm{,}
$$
while the "complex-conjugation" operator $K$ acts on complex numbers $\alpha\in\mathbb{C}$ as
$$
K\alpha = \alpha^* \rm{.}
$$

Now, given this information, we investigate the properties of an anti-unitary operator $A=UK$:
$$
\begin{align*}
&\langle A\phi, A\psi \rangle = \langle UK\phi, UK\psi \rangle = \langle K\phi, U^\dagger UK\psi \rangle = \langle K\phi, K\psi \rangle = \\
&= \sum_{i} \left(K\phi_i\right)^* \left(K\psi_i\right) = \sum_{i} \left(\phi_i^*\right)^* \psi_i^* = \sum_{i} \phi_i \psi_i^* = \left(\sum_{i} \phi_i^* \psi_i\right)^* = \langle \phi, \psi \rangle^* \rm{,}
\end{align*}
$$
so you obtain Eq.(32), the main result, without inventing constructs such as "a bra multiplied by anti-linear operator, such that it acts on kets, but first we must make this action linear, so we impose complex conjugation, but now we need the Hermitian conjugate for anti-unitary operators, so now we must be careful about writing the parentheses correctly in appropriate places..." which can be found in this note.

Ultimately, everything boils down to the observation (motivated by conservation of probability) that
$$
|\langle T\phi, T\psi\rangle| = |\langle \phi, \psi\rangle|
$$
is satisfied both by unitary operators, $\langle T\phi, T\psi \rangle = \langle\phi,\psi\rangle$, and by anti-unitary operators, $\langle T\phi, T\psi\rangle = \langle \phi, \psi\rangle^*$, and it's a matter of your own preference if you want to motivate either of these definitions in an alternative way (one of them is given in this note, for example; here, I gave another option).

 

[hokhani]

your problem is you need to apply the definition to figure out what $(<\phi|\hat{A})$ is.
$$<\phi|\hat{A} = -i\phi_2<+| + i\phi_1<-|$$

Thanks, could you please explain more how did you get this equation? I considered $|\alpha \rangle =\hat{A}^{\dagger} |\phi \rangle =K^{\dagger} L^{\dagger}=-i\phi_2^*|+\rangle +i \phi_1^* |-\rangle$ (supposing naively $K^{\dagger}=K$), then we have $\langle \alpha | =\langle \phi| \hat{A}=i\phi_2 \langle+| -i\phi_1 \langle -|$. So, I think my mistake is the calculation of $K^{\dagger}$ which I don't know how to calculate.

 

[QuarkyMeson]

Thanks, could you please explain more how did you get this equation? I considered $|\alpha \rangle =\hat{A}^{\dagger} |\phi \rangle =K^{\dagger} L^{\dagger}=-i\phi_2^*|+\rangle +i \phi_1^* |-\rangle$ (supposing naively $K^{\dagger}=K$), then we have $\langle \alpha | =\langle \phi| \hat{A}=i\phi_2 \langle+| -i\phi_1 \langle -|$. So, I think my mistake is the calculation of $K^{\dagger}$ which I don't know how to calculate.

Yes, it's going to seem rather silly I think when you see it:

Write the general form of $(<\phi|\hat{A})|\psi> = a\psi_1 + b\psi_2$

Now use your given definition to figure out a and b, $(a\psi_1 + b\psi_2) = [<\phi|(\hat{A}|\psi>)]^* = i\phi_1\psi_2-i\phi_2\psi_1$

So $b = i\phi_1$ and $a = -i\phi_2$

So the general construction for the object $<\phi|\hat{A} = -i\phi_2<+| + i\phi_1<-|$

Nothing fancy, it's just using the given definition.

 

[hokhani]

Let's calculate as follows:
$|\alpha \rangle =\hat{A}^{\dagger} |\phi\rangle =K^{\dagger} L^{\dagger}=K^{\dagger} \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix} \begin{bmatrix} \phi_1 \\ \phi_2 \end{bmatrix}=K^{\dagger} \begin{bmatrix} i\phi_2 \\ -i\phi_1 \end{bmatrix}$
so,
$\langle \alpha |=\langle \phi | \hat{A}=\begin{bmatrix} -i\phi_2^* & i\phi_1^* \end{bmatrix} K$
and
$(\langle \phi |\hat{A}) |\psi\rangle=\langle \alpha|\psi \rangle= \begin{bmatrix} -i\phi_2^* & i\phi_1^* \end{bmatrix} K \begin{bmatrix} \psi_1 \\ \psi_2 \end{bmatrix}=\begin{bmatrix} -i\phi_2^* & i\phi_1^* \end{bmatrix} \begin{bmatrix} \psi_1^* \\ \psi_2^* \end{bmatrix}=-i\phi_2^* \psi_1^*+i\phi_1^* \psi_2^*$
while
$<\phi|(\hat{A}|\psi>)^* = i\phi_1\psi_2-i\phi_2\psi_1$
which are not at all the same although this calculation seems correct.

 

[QuarkyMeson]

If you keep reading through the link you've posted you'll see that you're defining $A^{\dagger}$ wrong. $A^{\dagger} = KL^{\dagger}$ . You can reach this same conclusion using the definitions as above. Aka K is antiunitary.

 

[hokhani]

$A^{\dagger} = KL^{\dagger}$ . You can reach this same conclusion using the definitions as above.

Thanks, but doing calculation considering $A^{\dagger} = KL^{\dagger}$ we have:
$|\alpha \rangle =\hat{A}^{\dagger} |\phi\rangle =K L^{\dagger}=K \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix} \begin{bmatrix} \phi_1 \\ \phi_2 \end{bmatrix}=K\begin{bmatrix} i\phi_2 \\ -i\phi_1 \end{bmatrix}=\begin{bmatrix} -i\phi_2^* \\ i\phi_1^* \end{bmatrix}$
and
$(\langle \phi |\hat{A}) |\psi\rangle=\langle \alpha| \psi \rangle=i\phi_2 \psi_1 - i\phi_1 \psi_2$ which again results in $(\langle \phi |\hat{A}) |\psi\rangle=- <\phi|(\hat{A}|\psi>)^*$ as before!

Aside that, when $A=LK$, the relation $A^{\dagger} = KL^{\dagger}$ seems not correct in the algebra of operations.
I would appreciate any help.

 

[QuarkyMeson]

Did you read section 8 and 9 in the attached notes? What parts aren't clear?

 

[hokhani]

Did you read section 8 and 9 in the attached notes? What parts aren't clear?

Everything is ok if we have $(LK)^{\dagger}=L^{\dagger} K$. Is this equation correct?

 

[div_grad]

Everything is ok if we have $(LK)^{\dagger}=L^{\dagger} K$. Is this equation correct?

In general, if the operator $C = A B$ is made out of two operators $A$ and $B$, then from
$$
\langle \psi, C \psi \rangle = \langle C^\dagger \psi, \psi \rangle
$$
and
$$
\langle \psi, (AB) \psi \rangle = \langle A^\dagger \psi, B\psi \rangle = \langle B^\dagger A^\dagger\psi, \psi \rangle
$$
you get
$$
C^\dagger = (AB)^\dagger = B^\dagger A^\dagger \rm{,}
$$
which is the same as in linear algebra for matrices: $(M_1 M_2)^\dagger = M_2^\dagger M_1^\dagger$. So if you want to treat $K$ as an operator, then it should be that $(LK)^\dagger = K^\dagger L^\dagger$. You must figure out what is $K^\dagger$ then, which is why in post #3 I used a different method than the author of the note you posted.

 

[hokhani]

I used a different method than the author of the note you posted.

Thanks: right, your method was very nice, but I would like to try this method which still I am stuck in that.