Schrodinger Equation from Ritz Variational Method

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SUMMARY

The discussion focuses on the application of the Ritz Variational Method as outlined in W. Greiner's "Quantum Mechanics." Participants clarify that the derivatives in equations 11.35a and 11.35b are functional derivatives, allowing differentiation under the integral sign. This leads to the conclusion that the integral expressions simplify to ##\hat{H}\psi## and ##\psi## respectively. Additionally, the variation of the functional ##\bra{\psi}\hat{H}\ket{\psi}## is confirmed to yield ##\hat{H}\psi \delta \psi^{*}##, emphasizing the independence of the functional derivative with respect to ##\psi## and ##\hat{H}##.

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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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