I Schrodinger Equation from Ritz Variational Method

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The discussion clarifies that the derivatives in equations 11.35a and 11.35b are functional derivatives, allowing differentiation under the integral sign. This leads to the simplification where ##\int d^3x (\hat{H}\psi)## is treated as ##\hat{H}\psi## and ##\int d^3x \psi## as ##\psi##. Additionally, the replacement of ##\delta \bra{\psi}\hat{H}\ket{\psi}## with ##\hat{H}\psi## and ##\delta \bra{\psi}\ket{\psi}## with ##\psi## is justified because the functional derivative with respect to ##\psi^{*}## does not affect ##\psi## or ##\hat{H}##. This understanding is crucial for applying the Ritz Variational Method effectively in quantum mechanics. The conversation emphasizes the importance of recognizing the nature of functional derivatives in these equations.
Samama Fahim
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(This is from W. Greiner Quantum Mechanics, p. 293 from the topic of Ritz Variational Method)

1) Are ##\frac{\delta}{\delta \psi^{*}}## derivatives in equations 11.35a and 11.35b? If this is so, we can differentiate under the integral sign to get ##\int d^3x (\hat{H}\psi)## in equation 11.35a and ##\int d^3x \psi## in 11.35b, but why would ##\int d^3x (\hat{H}\psi)## be equal to just ##\hat{H}\psi## and ##\int d^3x \psi## to just ##\psi##?

2) Moreover, we replace ##\delta \bra{\psi}\hat{H}\ket{\psi}## in 11.34 with ##\hat{H}\psi## and ##\delta \bra{\psi}\ket{\psi}## with ##\psi## resulting in the equation at the bottom. Why is that? Is it that ##\delta (\psi^{*}\hat{H}\psi) = \hat{H}\psi \delta \psi^{*}## because we are looking at variation in ##\psi^{*}## and so we can take ##\hat{H} \psi## out?
 
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1) Those are functional derivatives, not regular derivatives.

2) The functional derivative wrt ##\psi^*## does not vary ##\psi## or ##H##.
 
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