What happened to the denominator?

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The denominator is 1 if all the kets are normalized, and they're saying at the bottom of 1-20 that they are.
 
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Ahhh. Thanks, I missed that.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
Thread 'Lesser Green's function'
The lesser Green's function is defined as: $$G^{<}(t,t')=i\langle C_{\nu}^{\dagger}(t')C_{\nu}(t)\rangle=i\bra{n}C_{\nu}^{\dagger}(t')C_{\nu}(t)\ket{n}$$ where ##\ket{n}## is the many particle ground state. $$G^{<}(t,t')=i\bra{n}e^{iHt'}C_{\nu}^{\dagger}(0)e^{-iHt'}e^{iHt}C_{\nu}(0)e^{-iHt}\ket{n}$$ First consider the case t <t' Define, $$\ket{\alpha}=e^{-iH(t'-t)}C_{\nu}(0)e^{-iHt}\ket{n}$$ $$\ket{\beta}=C_{\nu}(0)e^{-iHt'}\ket{n}$$ $$G^{<}(t,t')=i\bra{\beta}\ket{\alpha}$$ ##\ket{\alpha}##...

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