I cannot make the Gram-Schmidt procedure work

[George Keeling]

Homework Statement

I have been asked to use the Gram-Schmidt procedure and it does not produce orthonormal vectors :(

Relevant Equations

Gram-Schmidt procedure, inner product of complex vectors

This is problem A.4 from Quantum Mechanics – by Griffiths & Schroeter.

I cannot make the Gram-Schmidt procedure work. I don't know whether I am just inept with complex vectors or I have made some wrong assumption.

The Gram-Schmidt procedure (modified, I think)
Suppose you start with a basis $\left(|\left.e_1\right>,|\left.e_2\right>,\ldots,|\left.e_n\right>\right)$ that is not orthonormal. The Gram-Schmidt procedure is a systematic ritual for generating from it an orthonormal basis $\left(|\left.{e^\prime}_1\right>,|\left.\left.{e^\prime}_2\right>\right>,\ldots,|\left.{e^\prime}_n\right>\right)$. It goes like this:
(i) Normalize the first basis (divide it by its norm):
\begin{align}|\left.{e^\prime}_1\right>=\frac{|\left.e_1\right>}{\left|\left.e_1\right.\right|}&\phantom {10000}(1)\nonumber\end{align} (ii) Find the projection of the second vector along the first and subtract it off:
\begin{align}|\left.e_2\right>-\left<{e^\prime}_1|e_2\right>|\left.{e^\prime}_1\right>&\phantom {10000}(2)\nonumber\end{align}This vector is orthogonal to $|\left.{e^\prime}_1\right>$; normalise it to get $|\left.{e^\prime}_2\right>$.
(iii) Subtract from $|\left.e_3\right>$ its projections along $|\left.{e^\prime}_1\right>$ and $|\left.{e^\prime}_2\right>$ and get
\begin{align}|\left.e_3\right>-\left<{e^\prime}_1|e_3\right>|\left.{e^\prime}_1\right>-\left<{e^\prime}_2|e_3\right>|\left.{e^\prime}_2\right>&\phantom {10000}(3)\nonumber\end{align}This vector is orthogonal to $|\left.{e^\prime}_1\right>$ and $|\left.{e^\prime}_2\right>$; normalise it to get $|\left.{e^\prime}_3\right>$. And so on.

The authors also uses the * sign. For example at A.24 they almost say "with an orthonormal basis the inner product of two vectors can be written very neatly in terms of their components
\begin{align}\left<\beta|\alpha\right>={\beta_1}^\ast\alpha_1+{\beta_2}^\ast\alpha_2+\ldots+{\beta_n}^\ast\alpha_n&\phantom {10000}(4)\nonumber\end{align}". I assumed that ${\beta_1}^\ast$ is the complex conjugate of $\beta_1$. (Wikipedia agrees that physicists do this). Formula (4) will be very useful for executing the GS procedure.

The Question and my attempted solution
Use the Gram-Schmidt procedure to orthonormalize the 3-space basis
\begin{align}|\left.e_1\right>&=\left(1+i\right)\hat{i}+\hat{j}+i\hat{k}&\phantom {10000}(5)\nonumber\\|\left.e_2\right>&=i\hat{i}+3\hat{j}+\hat{k}&\phantom {10000}(6)\nonumber\\|\left.e_3\right>&=28\hat{j}&\phantom {10000}(7)\nonumber\end{align}The author does not say so but I assumed that $\hat{i},\hat{j},\hat{k}$ are an orthonormal basis. Although this begs the question: Why do we need another orthonormal basis?

It's obviously easiest to start with $|\left.e_3\right>$ and then it turns out that $|\left.e_2\right>$ is next easiest. Here they are:
\begin{align}|\left.{e^\prime}_3\right>=\hat{j}&\phantom {10000}(8)\nonumber\\|\left.{e^\prime}_2\right>=\frac{i\hat{i}+\hat{k}}{\sqrt2}&\phantom {10000}(9)\nonumber\end{align}$|\left.{e^\prime}_2\right>$ has norm 1 and is orthogonal to $|\left.{e^\prime}_3\right>$. Those were quite easy but this is where it all goes wrong: The third step is to find a vector orthogonal to $|\left.{e^\prime}_3\right>,|\left.{e^\prime}_2\right>$ which is
\begin{align}|\left.e_1\right>-\left<{e^\prime}_3|e_1\right>|\left.{e^\prime}_3\right>-\left<{e^\prime}_2|e_1\right>|\left.{e^\prime}_2\right>&\phantom {10000}(10)\nonumber\\=\left(1+i\right)\hat{i}+\hat{j}+i\hat{k}-\left(1\right)\hat{j}-\left(\left(\frac{i\hat{i}+\hat{k}}{\sqrt2}\right)\bullet\left(\left(1+i\right)\hat{i}+\hat{j}+i\hat{k}\right)\right)\frac{i\hat{i}+\hat{k}}{\sqrt2}&\phantom {10000}(11)\nonumber\\=\left(1+i\right)\hat{i}+i\hat{k}-\frac{1}{2}\left(1\right)\left(i\hat{i}+\hat{k}\right)=\left(1-\frac{i}{2}\right)\hat{i}+\left(-\frac{1}{2}+i\right)\hat{k}&\phantom {10000}(12)\nonumber\end{align}I have used $\bullet$ to indicate inner product when the first vector's components still need to be 'complex conjugated'. The vector has no $\hat{j}$ component so it is orthogonal to $|\left.{e^\prime}_3\right>$. Now normalise that and we have
\begin{align}|\left.{e^\prime}_1\right>=\frac{\frac{1}{2}\left(\left(2-i\right)\hat{i}+\left(-1+2i\right)\hat{k}\right)}{\sqrt{\left(\frac{1}{2}\left(\left(2-i\right)\hat{i}+\left(-1+2i\right)\hat{k}\right)\right)\bullet\left(\frac{1}{2}\left(\left(2-i\right)\hat{i}+\left(-1+2i\right)\hat{k}\right)\right)}}&\phantom {10000}(13)\nonumber\\|\left.{e^\prime}_1\right>=\frac{\left(2-i\right)\hat{i}+\left(-1+2i\right)\hat{k}}{\sqrt{10}}&\phantom {10000}(14)\nonumber\end{align}Now we just need to check that $|\left.{e^\prime}_1\right>,|\left.{e^\prime}_2\right>$ are orthogonal:
\begin{align}\left<{e^\prime}_1|\ {e\prime}_2\right>=\frac{1}{\sqrt{20}}\left(\left(2-i\right)\hat{i}+\left(-1+2i\right)\hat{k}\right)\bullet\left(i\hat{i}+\hat{k}\right)=\frac{-2}{\sqrt{20}}&\phantom {10000}(15)\nonumber\end{align}$|\left.{e^\prime}_1\right>,|\left.{e^\prime}_2\right>$ are not orthogonal! Where I have gone wrong?

 

[hutchphd]

Homework Statement: I have been asked to use the Gram-Schmidt procedure and it does not produce orthonormal vectors :(
Relevant Equations: Gram-Schmidt procedure, inner product of complex vectors

". I assumed that β1∗ is the complex conjugate of β1. (Wikipedia agrees that physicists do this). Formula (4) will be very useful for executing the GS procedure.

This is correct. So then $$\langle e'_1|= ???? $$

 

[George Keeling]

The same as $|\left.{e^\prime}_1\right>$ but using the complex conjugate of each component?

Because $\langle e'_1|e'_1\rangle|$ is the norm (squared).

 

[hutchphd]

Did you do that? I don't think so.

 

[George Keeling]

Did I do what? Sorry I dont understand

 

[hutchphd]

Eq 13 is correct but equation 15 is not. bras and kets

 

[vela]

Eq 13 is correct but equation 15 is not. bras and kets

I think (13) is wrong as well because of a sign error. The $\hat i$ component should $(2+i)/\sqrt{10}$.

 

[hutchphd]

Very possible but the big error is how to write the $\langle bra|ket \rangle$

 

[George Keeling]

Thanks everybody. I will return to the fray tomorrow.

 

[FactChecker]

I want to comment on how well this question was presented. The equation numbers and good formatting make it a real pleasure to discuss. I have not looked at the details, but the problem with the $\hat {i}$ component of equation (12) is clear.

 

[vela]

Very possible but the big error is how to write the $\langle bra|ket \rangle$

I think you missed that the OP wrote:

I have used ∙ to indicate inner product when the first vector's components still need to be 'complex conjugated'.

 

[hutchphd]

I think you missed that the OP wrote:

Yes I did miss that (very odd) notation. I believe the OP just screws up the final algebra step in line 15. I get zero.....but I do algebra badly. Whew!

 

[Steve4Physics]

I found checking the OP’s working awkward so would like to add this…

Using suitable notation can help minimise errors, improve readability and reduce eye-strain! It’s tempting here to:

- express the vectors in, say, row/column notation (common practice) to eliminate the need for $\hat i, \hat j$ and $\hat k$; this (IMO) makes things visually cleaner (as well as reducing the risk of getting $\hat i$ and $i$ muddled);

- use, say, $b_1, b_2$ and $b_3$ rather than $e'_1, e'_2$ and $e'_3$ as the new basis vectors to avoid clutter/confusion between "$e$"s and "$e’$"s.

The final answer can be translated back to the original format, if required.

I know many will already be familiar with this, but for the OP's benefit...

We can write:
$|e_1> = [1+i ~~1~~i]^T,~~|e_2> =[i~~3~~1]^T,~~|$ and $|e_3>=[0~~28~~0]^T$

From the OP's working we have:
$|b_3> = [0~~1~~0]^T$ and $|b_2> = \frac 1{\sqrt 2}[i~~0~~1]^T$.

We express $<b_3|$ and $<b_2|$ as row vectors and incorporate complex conjugation:
$<b_3| = [0~~1~~0]$ and $<b_2| = \frac 1{\sqrt 2}[-i~~0~~1]$

Our last basis vector (not normalised) is then:

$|b_{1,unorm }> = |e_1> - <b_3|e_1>|b_3> - <b_2|e_1>|b_2>$

$= [1+i ~~1~~i]^T - ([0~~1~~0][1+i ~~1~~i]^T)[0~~1~~0]^T - (\frac 1{\sqrt 2}[-i~~0~~1][1+i~~1~~i]^T)\frac 1{\sqrt 2}[i~~0~~1]^T$

$= [1+i~~1~~i]^T - (0.(1+i)+1.1+0.i)[0~~1~~0]^T - \frac 1{\sqrt 2}((-i)(1+i) + 0.1+ 1.i)\frac 1{\sqrt 2}[i~~0~~1]^T$

etc.

Even if there are mistakes, the above format should make locating them a bit easier.

 

[hutchphd]

I would also point out that one can write down <b3| by inspection after you see the form of the the other two b vectors. The Gram -Schmidt is really cumbersome. So $$\langle b_1 |=\frac 1 {\sqrt 2} [~ i~0~1~ ]$$

Homework Statement: I have been asked to use the Gram-Schmidt procedure and it does not produce orthonormal vectors :(
Relevant Equations: Gram-Schmidt procedure, inner product of complex vectors

Why do we need another orthonormal basis?

Bases are chosen typically using symmetries of the problem to male the arithmetic less opaque.

 

[George Keeling]

I think (13) is wrong as well because of a sign error. The $\hat i$ component should $(2+i)/\sqrt{10}$.

That has the great benefit of making $|\left.{e^\prime}_1\right>,|\left.{e^\prime}_2\right>$ orthogonal. Going back to (12) and extracting the $\hat{i}$
component. I wrote
$$\left(1+i\right)\hat{i}-\frac{1}{2}\left(1\right)\left(i\hat{i}\right)=\left(1-\frac{i}{2}\right)\hat{i}$$
and that is wrong.

It should have been
$$\left(1+i\right)\hat{i}-\frac{1}{2}\left(1\right)\left(i\hat{i}\right)=\left(1+i\right)\hat{i}-\frac{1}{2}i\hat{i}=\left(1+\frac{i}{2}\right)\hat{i}$$So it was just my ineptitude at complex vector arithmetic.
@vela Did you spot that because you knew the vector that would be orthogonal or because you saw my arithmetic error? Or something else?

Thanks again everybody!

 

[vela]

Did you spot that because you knew the vector that would be orthogonal or because you saw my arithmetic error?

I had Mathematica do the calculations you did and compared its results to yours. I didn't do the calculations by hand because I knew I'd probably make a mistake somewhere along the way too.

Sign errors are really easy to make, and it only gets worse when you throw $i$'s into the mix. It's one of the first things I look for if I'm expecting an answer of 0 but don't get it.

 

[WWGD]

I don't know if it would be over the top to suggest that a specific inner-product be specified. In this case, it's the "standard" one in $\mathbb R^n$ (n=2) here.