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Hermitian conjugate of outer product
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[QUOTE="loginorsinup, post: 4978643, member: 527065"] That's interesting. So basically everything in quantum mechanics involves operators since even states are operators? I'm a little confused about the terminology. I thought column vectors (kets) and row vectors (bras) were independent entities -- separate from matrices (operators). But, I guess you could say a column vector is simply a [itex]1\times n[/itex] matrix and a row vector is simply a [itex]n\times 1[/itex] matrix. You also noted this, so it seems like this is the case. Since [itex]\left<\alpha|\psi\right>[/itex] is just a complex-valued constant, I could say that [itex]\left|\beta\right>\left<\alpha|\psi\right>=c_{\beta}\left|\beta\right>[/itex]. It's corresponding ket would be [itex]c_{\beta}^*\left<\beta\right|=\left(\left<\alpha|\psi\right>\right)^*\left<\beta\right|=\left(\left<\psi|\alpha\right>\right)\left<\beta\right|[/itex] I have recently read that it is best to work with operators with some kind of trial function or bra or ket, which is what you suggested with that massive hint that I very much appreciate. So since the dual correspondence principle was assumed to be true, and I used it to find this relation that the dual of [itex]\left|\beta\right>\left<\alpha\right|[/itex] is [itex]\left|\alpha\right>\left<\beta\right|[/itex], I'm done. Thank you very much for your help. (I looked at your profile. I noticed you're grading exams. So, extra thanks for taking time out of your schedule to lend me a hand. :) ) [/QUOTE]
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Hermitian conjugate of outer product
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