If you mean generalized eigenvalue problem for non hermitian matrices NO. The Hamiltonian is Hermitian.<br><br>cheers<br><br>Layla<br><br><div class="gmail_quote">2012/10/3 Padmaja Patnaik <span dir="ltr"><<a href="mailto:padmaja_patnaik@yahoo.co.uk" target="_blank">padmaja_patnaik@yahoo.co.uk</a>></span><br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div><div style="font-size:12pt;font-family:times new roman,new york,times,serif"><div><span>Hi All</span></div><div style="font-style:normal;font-size:16px;background-color:transparent;font-family:times new roman,new york,times,serif">
<br><span></span></div><div style="font-style:normal;font-size:16px;background-color:transparent;font-family:times new roman,new york,times,serif"><span>I have a query. Anybody please can explain this. If you use the plane wave basis, will you have generalized  eigenvalue problem?</span></div>
<span class="HOEnZb"><font color="#888888"><div> </div><div>Padmaja Patnaik<br>Research Scholar<br>Dept of Physics<br>IIT Bombay<br>Mumbai, India</div></font></span></div></div><br>_______________________________________________<br>

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