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From the definition of spectral calculus, I want to verify whether the following operator identity is actually correct:

$$ (-1)^{\frac{1 + \sigma^z}{2}} = \sigma^z $$

This expression involves raising a number $-1$ to an operator exponent — which is not generally defined in operator algebra — unless we interpret it via functional calculus, specifically the spectral theorem.

Let me outline what I believe is the intended meaning.

We define the function $$ f(\sigma^z) := (-1)^{\frac{1 + \sigma^z}{2}} $$ using the spectral theorem. That is, for any state $|\psi\rangle$ that is an eigenstate of $\sigma^z$, with eigenvalue $\lambda \in \{+1, -1\}$, we define: $$ f(\sigma^z) |\psi\rangle = f(\lambda) |\psi\rangle. $$

So:

  • $f(+1) = (-1)^1 = -1$
  • $f(-1) = (-1)^0 = +1$

Hence, $$ (-1)^{\frac{1 + \sigma^z}{2}} = -|\uparrow\rangle \langle \uparrow| + |\downarrow\rangle \langle \downarrow| = \sigma^z. $$

My questions are:

  • Is this identity indeed correct under the rules of spectral calculus?
  • Is this a valid application of the spectral theorem even though the function $f(x) = (-1)^x$ is not analytic (i.e., not expandable as a power series)?
  • Are there any subtleties in applying spectral calculus to such functions — particularly when they’re defined only on the spectrum of the operator?

I'm looking for a mathematically precise explanation — either confirming the identity or pointing out any caveats I should be aware of.

Qmechanic
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felix
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1 Answers1

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You have a mistake somewhere. You need a minus sign. Namely, $$\tag1\text{either}\qquad\left(-1\right)^\dfrac{1+\sigma^z}2=-\sigma^z\qquad\text{or}\qquad\left(-1\right)^\dfrac{1-\sigma^z}2=\sigma^z$$ The correct way to deal with this is Borel functional calculus and you are almost there anyway.

You could also write $-1=\mathrm e^{\pm\mathrm i\pi}$ and then you can use the standard Lie group Lie algebra stuff, or even the standard Physics hack of equating things with their Taylor series expansion.

naturallyInconsistent
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