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</div><div id="yui_3_16_0_ym18_1_1459363378947_2643"><span id="yui_3_16_0_ym18_1_1459363378947_2644" style="font-size:11.0pt">Chandra:</span></div>
<div id="yui_3_16_0_ym18_1_1459363378947_2645"><span id="yui_3_16_0_ym18_1_1459363378947_2646" style="font-size:11.0pt">Yes, there are others that
seem to use “standing waves” and the physics of standing waves in an
propagating wave scenario. This is a theory no-no in the sense it is a
redefinition of terms during the argument. </span></div>
<div id="yui_3_16_0_ym18_1_1459363378947_2647"><span id="yui_3_16_0_ym18_1_1459363378947_2648" style="font-size:11.0pt">Note that standing waves may
be obtained by having a single source and reflection back to the source. I
suggest this occurs in the Transnational Interpretation of QM. The reverse time
wave can be the reflected wave of a standing wave if the wave speed is high
enough. Thus, as you note, the $\psi$ and $\psi^*$ produce the observable. A
standing wave analysis can produce an observable, but it must to be a true
standing wave (common source) and not a propagating wave. </span></div>
<div id="yui_3_16_0_ym18_1_1459363378947_2649"><span id="yui_3_16_0_ym18_1_1459363378947_2650" style="font-size:11.0pt"> </span></div>
<div id="yui_3_16_0_ym18_1_1459363378947_2651"><span id="yui_3_16_0_ym18_1_1459363378947_2652" style="font-size:11.0pt">The question of how emitted
energy cannot be exhausted remains. There must be a means to recover the energy
or the process decays. </span></div>
<div id="yui_3_16_0_ym18_1_1459363378947_2653"><span id="yui_3_16_0_ym18_1_1459363378947_2654" style="font-size:11.0pt"> </span></div><div dir="ltr">
<span id="yui_3_16_0_ym18_1_1459363378947_2655" style="font-size:11.0pt;font-family:"Times New Roman";mso-fareast-font-family:"Times New Roman";mso-ansi-language:EN-US;mso-fareast-language:EN-US;mso-bidi-language:AR-SA">Hodge</span></div></div></body></html>