# Griffiths use of a linear transformation on basis vectors. Need help with derivation

In Griffiths' intro to Quantum Mechanics 2nd edition, in the appendix A.3 which is a review on linear algebra and matrixes (on page 441), he states that a linear transformation on a set of basis vectors can be expressed as follows:

\begin{align*} \hat{T}|e_1\rangle &= T_{11}|e_1\rangle + T_{21}|e_2\rangle+...+T_{n1}|e_n\rangle\\ \hat{T}|e_2\rangle &= T_{12}|e_1\rangle + T_{22}|e_2\rangle+...+T_{n2}|e_n\rangle\\ &...\\ \hat{T}|e_n\rangle &= T_{1n}|e_1\rangle + T_{2n}|e_2\rangle+...+T_{nn}|e_n\rangle\\ \end{align*} Where $\hat{T}$ is a linear transformation and $|e_i\rangle$ is the $i$th basis vector in the basis set.

Where exactly is Griffith getting this? I have never seen this in any previous linear algebra text where a transform on a basis is expressed in terms of the other basis vectors. The ONLY way I see this working is if the basis set is the standard basis set:

\begin{align*} |e_1\rangle=\left[\begin{array}{c} 1\\0\\\vdots\\0 \end{array}\right],&& |e_2\rangle=\left[\begin{array}{c} 0\\1\\\vdots\\0 \end{array}\right],&& ...,&& |e_n\rangle=\left[\begin{array}{c} 0\\0\\\vdots\\1 \end{array}\right] \end{align*}

However in Griffiths, he doesn't make this distinction. He doesn't even qualify if the basis is orthonormal. I have tried to derive it myself without using the standard basis in $\mathbb{R}^n$ and am getting nowhere and just don't see it. Anyone have any advice?

Let $\{|e_i\rangle: 1\leq i\leq n\}$ be a basis.

Since $|e_1\rangle$ is a vector, and $\hat{T}$ is a linear transformation then $\hat{T}|e_1\rangle$ is a vector. Since $\hat{T}|e_1\rangle$ is a vector and $\{|e_i\rangle: 1\leq i\leq n\}$ is a basis then $\hat{T}|e_1\rangle$ is uniquely expressed as a linear combination of the vectors in $\{|e_i\rangle: 1\leq i\leq n\}$ and so we call those coefficients $T_{11},$ $T_{21},$ ..., and $T_{n1}$ (we have to call them something). Thus $\hat{T}|e_1\rangle = T_{11}|e_1\rangle + T_{21}|e_2\rangle+...+T_{n1}|e_n\rangle$ and similarly $\hat{T}|e_2\rangle$ is a vector so $\hat{T}|e_2\rangle$ is uniquely expressed as a linear combination of the basis vectors in $\{|e_i\rangle: 1\leq i\leq n\}$ and so we call those coefficients $T_{12},$ $T_{22},$ ..., and $T_{n2}$ (we have to call them something). Thus $\hat{T}|e_1\rangle = T_{11}|e_1\rangle + T_{21}|e_2\rangle+...+T_{n1}|e_n\rangle$ and similarly we repeat this for any basis vector, and write its image as a linear combination of the basis vectors and give the result names according to the above convention and get:

\begin{align*} \hat{T}|e_1\rangle &= T_{11}|e_1\rangle + T_{21}|e_2\rangle+...+T_{n1}|e_n\rangle\\ \hat{T}|e_2\rangle &= T_{12}|e_1\rangle + T_{22}|e_2\rangle+...+T_{n2}|e_n\rangle\\ &...\\ \hat{T}|e_n\rangle &= T_{1n}|e_1\rangle + T_{2n}|e_2\rangle+...+T_{nn}|e_n\rangle\\ \end{align*}

Personally I might prefer the naming convention that allows us to say:

\begin{align*} \hat{T}|e_1\rangle &= T_{11}|e_1\rangle + T_{12}|e_2\rangle+...+T_{1n}|e_n\rangle\\ \hat{T}|e_2\rangle &= T_{21}|e_1\rangle + T_{22}|e_2\rangle+...+T_{2n}|e_n\rangle\\ &...\\ \hat{T}|e_n\rangle &= T_{n1}|e_1\rangle + T_{n1}|e_2\rangle+...+T_{nn}|e_n\rangle\\ \end{align*}

First get the idea of kets as some component in $\mathbb{R}^n$ out of your mind. kets are elements of a complex vector space, in this case a finite dimensional one. Yes the space will be isomorphic to $\mathbb{R}^n$, or rather $\mathbb{C}^n$, but by assuming they are $\mathbb{C}^n$, you may imbue them with properties of $\mathbb{C}^n$ that are not true of a general vector space.

A basis is defined as a maximally linearly independent set of elements of the space. Linearly independent means that you cannot make a linear combination of the vectors which is zero ie $$0= A_1|e_1\rangle + A_2|e_2\rangle+...+A_n|e_n\rangle\\$$ can happen only if all the $A_n$ are 0.

Maxmimal means that you cannot add any more vectors to the set and have it be linearly independent.

A consequence of linear independence is that any vector $|\alpha\rangle$ van be written as linear combination of the kets. Proof: Take $|\alpha\rangle$ . We know that we can write $$0|=A|\alpha\rangle+B_1|e_1\rangle + B_2|e_2\rangle+...+B_n|e_n\rangle\\$$ and all the $B_n$ are not zero. If we could not, then $|\alpha\rangle,|e_n\rangle$ would be linearly independent and $|e_n\rangle$ would not be maximal.

We know that A cannot be zero, because if it was then we would have a linear combination of $|e_n\rangle$ which is zero. So divide by A and make $C_n=-B_n/A$. We get that: $$|\alpha\rangle=C_1|e_1\rangle + C_2|e_2\rangle+...+C_n|e_n\rangle\\$$

IF $T$ is a linear transformation what is $\hat{T}|e_i\rangle$? Answer: it's a ket So we can write $$\hat{T}|e_m\rangle = t_1|e_1\rangle + t_2|e_2\rangle+...+t_n|e_n\rangle\\$$

So now just write $T_{mn}=t_n$ and you get your result.

• I also tried something similar and got to the last part. But how do you prove or justify $t_n=T_{nm}$ for ANY set of basis vectors? That is where I get stuck. Like I said I can easily see it if the kets have components in $\mathbb{R}^n$ AND the basis set is the standard basis set. But how do you shows this for ANY basis set especially one with kets that have components in $\mathbb{C}^n$? – Wikkyd Oct 31 '15 at 16:17
• @Wikkyd: The equation $T_{nm}=T_n$ is the definition of $T_{nm}$. I'm not sure what you think there is to be shown. – ACuriousMind Oct 31 '15 at 23:24