Gram Schmidt procedure, trouble finding inner product

[Jimmy000]

Homework Statement

Given basis |x>,|y>,|z> such that <x|x> = 2,<y|y> = 2,<z|z> = 3,<x|y> = i, <x|z> = i, and <y|z> = 2. Build an orthonormal basis|x'>,|y'>,|z'>. Each of the new basis vectors should be expressed in terms of the old ones multiplied by coefficients.

Homework Equations

|x'> = |x>/(<x|x>)^.5 (normalizing the original x ket)

|y'> = |y> - <x'|y> |x'>

The Attempt at a Solution

So far I have worked through that |x'> = (1/2)^.5|x>
Using this result I have moved through the calculation of |y'> and found it equals

|y> - (i/2) |x'>

I am now trying to normalize this expression, but I am not sure what my inner product on the denominator will look like for this normalization, but my best guess is
<y| - (i/2)<x'|y> - (i/2)|x'>

but that does not look legitimate to me, so I am stuck here.

How do I take the inner product of a bra and and ket, both of which are sums of two kets/bras?

 

[vela]

Homework Statement

Given basis |x>,|y>,|z> such that <x|x> = 2,<y|y> = 2,<z|z> = 3,<x|y> = i, <x|z> = i, and <y|z> = 2. Build an orthonormal basis|x'>,|y'>,|z'>. Each of the new basis vectors should be expressed in terms of the old ones multiplied by coefficients.

Homework Equations

|x'> = |x>/(<x|x>)^.5 (normalizing the original x ket)

|y'> = |y> - <x'|y> |x'>

The Attempt at a Solution

So far I have worked through that |x'> = (1/2)^.5|x>
Using this result I have moved through the calculation of |y'> and found it equals

|y> - (i/2) |x'>

I am now trying to normalize this expression, but I am not sure what my inner product on the denominator will look like for this normalization, but my best guess is
<y| - (i/2)<x'|y> - (i/2)|x'>

but that does not look legitimate to me, so I am stuck here.

How do I take the inner product of a bra and and ket, both of which are sums of two kets/bras?

Use parentheses. If you had a = b + c, you wouldn't say $a\times a = b + c\times b + c$, would you? That's effectively what you're doing right now, so it's not surprising it doesn't look right to you.

 

[Fred Wright]

In the "bra"-"ket" formalisim the bra is the complex conjugate of the ket. In your normalizing denominator you must take the inner-product of the vector with its complex conjugate.

 

[Jimmy000]

Thanks a million! I always mix up how kets and bras correlate to normal vector notation. I worked it through being mindful of parentheses and I got a much more logical answer