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For Quantum-mechanics on a Hilbert-space over the complex numbers, the usual scalar product of two states $\langle \phi | \psi \rangle$ and gives the transition amplitude between the two states. The absolute square of this quantity then gives the probability that a particular value associated with $|\phi \rangle$ can be measured when the system is in state $| \psi \rangle$.

However, when one constructs states over super-numbers (for example fermionic coherent states), those states do have supernumbers as coefficients, and thus the scalar product yields a super-number as well.

Can this super-numbers still be used as a transition-amplitude?

For example, in a 2 state-system: $$ |\theta \rangle = | 0 \rangle - \theta | 1 \rangle \\ $$ then $$ \langle 0 |\theta \rangle = 1 \\ \langle 1 | \theta \rangle = - \theta. $$ How would we proceed from here?

  • The absolute square would be $ \bar{\theta} \theta $, which is grassmann even - or would it be $\theta \theta = 0$?
  • If the square is zero, does that mean that fermionic coherent states essentially are overlapping with the vacuum state?
  • Is the concept of transition probabilities simply not defined for states over super numbers?
  • If so, could it in principle be defined in a consistent way?
Qmechanic
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Quantumwhisp
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1 Answers1

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In order for the Born rule of a wavefunction or an overlap to produce measurable physical probabilities $\in[0,1]$ of ordinary numbers, all supernumbers must first have been integrated out, cf. e.g. this & this Phys.SE posts.

Qmechanic
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  • I would have guessed that integrating the supernumbers out is a fix, but in the general setting I supposed in the question, there might not be a path integral, or any other reason to integrate out the supernumbers. Should I understand your answer as the statement that any other way of producing probabilities will lead to inconsistencies, so integrating out here is the only way to produce those probabilities as well? – Quantumwhisp Jul 05 '22 at 12:42
  • Any other way of producing consistent probabilities is equivalent to integrating the supernumbers out. – Qmechanic Jul 05 '22 at 12:55
  • To take this further, this also means a coherent states norm is not defined until you integrate over it? (In a way this is analogue to a position eigenstates norm not being defined until you integrate over it ?). – Quantumwhisp Jul 06 '22 at 10:13
  • The overlap of 2 fermionic coherent states is a Grassmann-even supernumber. – Qmechanic Jul 06 '22 at 15:22