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Are these concepts equivalent? And if not, which one implies the other one?

A transformation $\hat U$ is unitary when $\hat U^{-1} = \hat U^{\dagger}$.

A reversible transformation $\hat A$ admits an inverse, $\hat A^{-1}$, that's all.

Then, what is the implicative relation between the two properties?

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    $\begingroup$ This question has been asked, and answered, before. Refer to physics.stackexchange.com/q/51424 $\endgroup$
    – nir
    Commented Dec 6, 2014 at 4:44
  • $\begingroup$ @ nir: 1) I don't ask here about entropy. $\endgroup$
    – Sofia
    Commented Dec 6, 2014 at 10:26

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I think you have all the knowledge you need. The definitions are as you state, and Unitarity implies reversibility (all unitary transformations have an inverse) but the converse doesn't hold. To prove that they are not the same concept, you simply need to show that there is a reversible, but non unitary transformation. Enter exhibit A: the following linear homogeneous transformation:

$$\mathscr{L}:(\mathbb{C}^2,\,+)\to (\mathbb{C}^2,\,+);\;\;\mathscr{L}\left(\begin{array}{c}z_1\\z_2\end{array}\right) = \left(\begin{array}{cc}1&z\\0&1\end{array}\right)\left(\begin{array}{c}z_1\\z_2\end{array}\right)$$

All matrix groups comprise only invertible transformations. Most of them are not unitary transformations: the following is a strict set inclusion $SU(2)\subset SL(2,\,\mathbb{C})$.

Is this what you mean? (I'm having a little trouble gauging your level, as I have seen some good answers from you, so I am a little surprised at your question - not being critical - I lack all kinds of knowledge that I should have too).

Now if your question is specifically to do with quantum mechanics, then quantum states always need to have an $\mathcal{L^2}$ length of one, if you accept the Born interpretation of the quantum state as a probability amplitude vector. All possibilities must sum to unity probability: the system must be in some state and, to be a full specification of the quantum state, the state vector must exhaust possibilities. So any realistic transformation of a quantum state must preserve the length. Any reversible linear transformation between quantum states must be unitary. Indeed, one doesn't have to assume linearity if one assumes that transformations wrought on quantum states are compatible with Lorentz transformations on the physical system, then the transformation must be linear and unitary; this is part of what Wigner's theorem tells us (which I talk about more in my answer here).

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  • $\begingroup$ @Savanna: No, no! I don't accept Bohm's interpretation, I INVESTIGATE it. I hope that you know that it is at odds with relativity. $\endgroup$
    – Sofia
    Commented Dec 6, 2014 at 10:52
  • $\begingroup$ @Savanna: Aye! You are great! I checked now, indeed the inverse of the matrix you showed, is the one with -z instead of z. Now, please see my level: I am not at home with group theory, neither with QFT, and I have no time for them. Neither can I see in the near future, some time becoming available. Thus, if possible, would you reformulate the answer without groups? Please try! $\endgroup$
    – Sofia
    Commented Dec 6, 2014 at 11:08
  • $\begingroup$ @Sofia the groups are simply to show that there are many $2\times2$ matrices which are not unitary. $\endgroup$ Commented Dec 6, 2014 at 11:39

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