Quick Dirac Notation question - (maths)

[square_imp]

Got a question as follows:

<aaa/aab> where the slash is meant to be a straight line down and a is alpha and b is beta. Alpha and Beta are orthogonal.

In this case can you work out the bra-ket by doing each pair, i.e aa = 1, aa = 1, ab = 0 ? Giving an answer of zero? I am not really sure. Alpha and Beta are the two states for nucleon spin.

generally any help is much appreciated.

Square

 

[Psiko]

I'm not really sure I understand the problem. Firstly, a and b being orthogonal does not necessarily mean that <a | a> = 1. That would be orthonormal (if <a | b> = 0 as well.)

If |a> and |b> are states, then I'm guessing that |aab> means the tensor product of the three states |a>, |a> and |b>, for three different particles. In that case (using x to mean tensor product)

<aaa|aab> = (<a| x <a| x <a|)(|a> x |a> x |b>) = <a|a><a|a><a|b> = 0.

To make it clearer you could add suffices to the bras and kets (1,2,3 for each particle, then the 1 bras only act on the 1 kets etc.)

However, I'm not entirely sure about this.

 

[spp]

hmmmm, it has been a good while since I've done any QM, but off the top of my head I think you can think of the ket as the column vector, (a,a,a)^T and the bra as (a*,a*,b*) where the stars denote the complex conjugates. Then <aaa|aab> is just the inner product, so equal to aa* + aa* + a*b, and a*b will be zero. Note this may not be right, so definitely best to check with someone.

 

[square_imp]

<aaa|aab> = (<a| x <a| x <a|)(|a> x |a> x |b>) = <a|a><a|a><a|b> = 0.

To make it clearer you could add suffices to the bras and kets (1,2,3 for each particle, then the 1 bras only act on the 1 kets etc.)

However, I'm not entirely sure about this.

that was my main question about whether the above manipulation is valid, then the problem is fine.

thanks for the help.

 

[Squink]

If |a> and |b> are states, then I'm guessing that |aab> means the tensor product of the three states |a>, |a> and |b>, for three different particles. In that case (using x to mean tensor product)

My cat's name is Mittens

 

[PhillyDee]

hmmmm, it has been a good while since I've done any QM, but off the top of my head I think you can think of the ket as the column vector, (a,a,a)^T and the bra as (a*,a*,b*) where the stars denote the complex conjugates. Then <aaa|aab> is just the inner product, so equal to aa* + aa* + a*b, and a*b will be zero. Note this may not be right, so definitely best to check with someone.

*snigger*

 

[sniffy]

My cat's name is Mittens

Hahahaha