# I Complex conjugate and time reversal operator

1. Sep 11, 2016

### dyn

Hi. I'm confused about the action of the complex conjugate operator and time reversal operator on kets.
I know K(a |α > ) = a* K | α > but what is the action of K on | α > where K is the complex conjugation operator ? What is the action of the time reversal operator Θ on a ket , ie. what is Θ | α > ?
Thanks

2. Sep 11, 2016

### Nosebgr

I don't quite understand the question here, please let me know if your question is different than the one I'm about to answer.
The complex conjugate of a ket vector is defined to be the corresponding bra vector, therefore K|a> would be equal to <a|. Just take the complex conjugate and turn the column vector into a row vector.
The action of the time reversal operator would depend on the time dependence of the state. I can't say much more than that it would reverse the time evolution of the state represented by the ket vector - by definition.

3. Sep 11, 2016

### dyn

By complex conjugate operator I mean K where the time reversal operator Θ is given by Θ = UK and U is a unitary operator

4. Sep 11, 2016

### Truecrimson

The action of the complex conjugate operator $K$ is basis dependent. If you choose a basis containing $| \alpha \rangle$ (and not $e^{i \theta }|\alpha \rangle$ for any nontrivial $\theta$), then $K$ does nothing to $| \alpha \rangle$.

5. Sep 12, 2016

### dyn

Thanks. So the complex conjugate operator does nothing to kets ; what about the time reversal operator ? How does that affect kets ?

6. Sep 12, 2016

### vanhees71

A complex-conjugate operator makes only sense if defined for a fixed orthonormal basis. By definition it doesn't change these basis vectors.

There's no such problem with the general coordinate free definition of an antilinear operator. It just fulfills for any linear combination of vectors
$$\hat{A} (\lambda_1 |\psi_1 \rangle + \lambda_2 \psi_2 \rangle = \lambda_1^* \hat{A} |\psi_1 \rangle + \lambda_2^* \hat{A} |\psi_2 \rangle.$$
An antilinear operator is called antiunitary if it fulfills in addition
$$\langle \hat{A} \psi_1|\hat{A} \psi_2 \rangle=\langle \psi_1|\psi_2 \rangle^*=\langle \psi_2|\psi_1 \rangle.$$
Wigner's theorem tells you that a symmetry in quantum theory (acting in the space of states, i.e., a projective Hilbert space), which keeps the moduli of scalar products invariant, can be lifted either to a unitary or an antiunitary operator on Hilbert space.

For a very detailed proof, see

Gottfried, Quantum Mechanics vol. I, 1st edition

Unfortunately this very nice proof is spoiled in the 2nd edition :-(.

7. Sep 12, 2016

### dyn

Thanks. How does the time reversal operator affect kets in general ?
Can i also check I have the following correct where | + > is the eigenvalue for the Sz operator.
Θ e- iπ Sy | + > = - e+ iπ Sy Θ | + >
As Θ and Sy anticommute and Θ takes the complex conjugate but i'm not sure about this last point as the exponential is an operator not just a number ?

8. Sep 13, 2016

### dyn

Could someone please tell me what Θ e- i π Sy is when the order of the time reversal operator and the exponential operator are reversed. I know Θ and Sy anticommute so it picks up a negative sign but as Θ is antiunitary does it cause the i in the exponential to change sign ?

9. Sep 14, 2016

### vanhees71

Let $\alpha \in \mathbb{R}$. The exponential function is defined by its Taylor series
$$\exp(-\mathrm{i} \alpha S_y)=1+(-\mathrm{i} \alpha) S_y +\frac{(\mathrm{i} \alpha)^2 S_y^2}{2!} + \frac{(-\mathrm{i} \alpha)^3 S_y^3}{3!} + \cdots$$
Now $\Theta$ anticommutes with each factor $(-\mathrm{i} \alpha)$ and with each factor $S_y$. So bringing the time-reversal operator $\Theta$ from the very left to the very right of each of the terms you have always an even number of factors $(-1)$ and thus nothing happens, i.e., you have
$$\Theta \exp(-\mathrm{i} \alpha S_y)=\exp(-\mathrm{i} \alpha S_y) \Theta.$$
This must be so, because a rotation doesn't change due to time reversals, i.e., you must have
$$\Theta \exp(-\mathrm{i} \vec{\phi} \vec{S}) \Theta^{-1}=\exp(-\mathrm{i} \vec{\phi} \vec{S}),$$
because in the time-reversed world time translations flip sign while spatial translations stay the same.

The same group-theoretical arguments lead also to the demand that $\Theta$ must be antiunitary, i.e., time-reversal must be realized as an antiunitary transformation. The same holds for space-time translations, i.e., under time reversal you must have for the energy and momentum
$$\Theta \mathrm{i} H \Theta^{-1}=-\mathrm{i} H, \quad \Theta \mathrm{i} \vec{p} \Theta^{-1}=+\mathrm{i} \vec{p}. \qquad (*)$$
Now energy is special. In order to have a stable ground state of your system the Hamiltonian must be bounded from below. So if the time-reversal operator is supposed to exist for your quantum system with $H$ also $\Theta H \Theta^{-1}$ must bebounded from below. To get this consistent with the symmetry property of $\Theta$ it thus must be realized as anti-unitary operator, because only then you get from the symmetry property (*) that
$$\Theta H \Theta^{-1}=+H,$$
i.e., the correct property that in the time-reflected system $H$ stays bounded from below.

10. Sep 14, 2016

### dyn

Thanks for all your replies and especially vanhees71. That last post of yours was very helpful and cleared up a lot of points for me !

11. Sep 15, 2016

### DrDu

[QUOTE="vanhees71, post: 5566235, member: 260864" In order to have a stable ground state of your system the Hamiltonian must be bounded from below. [/QUOTE]

Just one remark to your very nice explanation. As far as I understood, the requirement that energy be bounded from below and not from above is rather a convention than a physical necessity. The stability of the states depends rather on their entropy than their energy. Now entropy would remain invariant if we both change the sign of E and T. At the time when Wigner pondered about time reversal invariance, negative temperatures were probably not an option, but nowadays negative temperatures are quite a common concept in spin systems where energy is bounded both from below and from above.