- #1

- 466

- 5

|e>=(1+i,1,i) (n-tuple representation, where i's are the imaginaries)

so the norm of this would then be the following?

||e||=$$\sqrt{<e|e>}$$=$$\sqrt{(1+i,1,i)\cdot(1+i,1,i)}$$=$$\sqrt{(1+2i+i^2)+1+i^2}$$=$$\sqrt{2i}$$

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- Thread starter iScience
- Start date

- #1

- 466

- 5

|e>=(1+i,1,i) (n-tuple representation, where i's are the imaginaries)

so the norm of this would then be the following?

||e||=$$\sqrt{<e|e>}$$=$$\sqrt{(1+i,1,i)\cdot(1+i,1,i)}$$=$$\sqrt{(1+2i+i^2)+1+i^2}$$=$$\sqrt{2i}$$

- #2

jedishrfu

Mentor

- 12,809

- 6,690

It seems that you can extract the sqrt(i) from the sqrt though:

http://www.math.toronto.edu/mathnet/questionCorner/rootofi.html

- #3

Dick

Science Advisor

Homework Helper

- 26,260

- 619

|e>=(1+i,1,i) (n-tuple representation, where i's are the imaginaries)

so the norm of this would then be the following?

||e||=$$\sqrt{<e|e>}$$=$$\sqrt{(1+i,1,i)\cdot(1+i,1,i)}$$=$$\sqrt{(1+2i+i^2)+1+i^2}$$=$$\sqrt{2i}$$

No, it's not right. If |e>=(1+i,1,i) then <e| is the hermitian conjugate vector. You forgot the complex conjugation. <e|e> should be a real number.

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