- #1
George Keeling
Gold Member
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- 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?
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?
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