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I should prove that $$\exp \left ( \frac{\pi}{2\hbar^2}(L_x^2 + L_y^2) \right )$$ is not a unitary operator. Where $L$ is the total angular momentum of a 2-particle system ($L = L_A + L_B$ for the particles $A$ and $B$).

My (undergraduate) definition of unitary operator is:

$U$ is a unitary operator if $UU^+ = U^+U=I$, where $I$ is the identity operator and $U^+$ is the adjoint of $U$

I have tried using that $$e^U = \sum_k^{\infty}\frac{U^k}{k!}$$ but without any success. I also wrote down some properties of exponential matrices but I really don't know how to proceed here.

EDIT: I have noticed there is a typo in a subsequent part of the exercise ($L_z^2$ swapped with $L_z$). Could it be possible that I should consider $$\exp \left ( \frac{\pi}{2\hbar}(L_x + L_y) \right )?$$

EDIT2: It was $\hbar^2$ and not $\hbar$. And I posted an answer.

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    $\begingroup$ Are your $L_i$ Hermitian or anti-Hermitian, and is there really no $\mathrm{i}$ in the exponential or did you forget it? $\endgroup$ – ACuriousMind Feb 3 '19 at 14:44
  • $\begingroup$ @ACuriousMind There is no $i$ in the exponential. $L$ is Hermitian and I have used the basis $|L L_z \rangle$ for previous questions of the exercise. So I am quite sure $L_i$ are Hermitian (they are observables) $\endgroup$ – moonknight Feb 3 '19 at 14:52
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How about this approach: Write the operator as $\exp \left ( \frac{\pi}{2\hbar}(L^2 - L_z^2) \right )$. Then consider what happens when you apply it to an angular momentum eigenstate $\left|l\,m \right>$. Does it change the norm of the state? If so then it can't be unitary.

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  • $\begingroup$ This is a great answer! $\endgroup$ – secavara Feb 3 '19 at 15:35
  • $\begingroup$ I have just posted my answer, I think it should okay and I only used linearity and properties of angular momentum. $\endgroup$ – moonknight Feb 3 '19 at 15:40
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Try using the following facts:

  1. Adjunction ${}^\dagger$ is linear as $(A+B)^\dagger = A^\dagger + B^\dagger$, so $\exp(U)^\dagger = \exp(U^\dagger)$.

  2. Any operator that is both Hermitian and unitary is a projection (cf. this math.SE post), i.e. has eigenvalues $0,1$.

  3. $O(t) = \exp(\frac{t\pi}{2h}(L_x^2 + L_z^2))$ on $[0,1]$ smoothly interpolates between the identity and your operator. The only projector which you can smoothly obtain from the identity is the identity itself.

Note that none of these things is specific to what the argument of the exponential actually is.

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  • $\begingroup$ During the Quantum Mechanics course we didn't talk much about operators, so I am not supposed to know about the second (2.) fact. It is a great way to prove it but I think I have just found another way a bit simpler. I'll post the answer. $\endgroup$ – moonknight Feb 3 '19 at 15:17
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I think I got it: $$A=\exp \left ( \frac{\pi}{2\hbar^2}(L_x^2+L_y^2) \right ) $$

Now $L_x^2$ and $L_y^2$ are Hermitian operators (they are because $(L_x^2)^+ = (L_xL_x)^+=L_x^+L_x^+=L_x^2$ using the fact that $L_x$ is hermitian) and using $\exp(U)^+ = \exp(U^+)$ I have: $$A^+ = A$$ Then $AA^+=A^2 \not= I$.

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    $\begingroup$ Strictly speaking, you still have to prove that $A^2 \neq I$. $\endgroup$ – Javier Feb 3 '19 at 15:52
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Suppose $\hat A$ is an Hermitian operator. This means all its eigenvalues are real. Exponentiation of operator amounts to exponentiation of each of its eigenvalues. Since your exponentiation expression doesn't contain $i$, and eigenvalues of $\hat A$ are not all zeros, eigenvalues of $\exp(\hat A)$ don't all have absolute value $1$. Thus the exponentiated operator can't be unitary.

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