I have a problem at understanding the way linear transformations are used in Griffiths Introduction to Quantum Mechanics. My knowledge about linear algebra is basic. He is making a reference to the linear transformations of the basis vectors so I would expect that he implies the multiplication of a linear transformation $\hat T$ with the basis vectors which are set in a column (The transformation is located to the left while the vector is located to right during multiplication). However, the result of the multiplication is not what I expect and it seems to me that he multiplied the basis vectors that are set in a row with the row being located to the left, during multiplication, with the linear transformation which is located to the right. Is my understanding of what he is doing correct? If yes, isn't the transformation supposed to be located to the left?

The relevant passage is reproduced below.

3.1.3 Linear Transformations

Suppose you take every vector (in three-space) and multiply it by 17, or you rotate every vector by 39$^\circ$ about the z-axis, or you reflect every vector in the $xy$-plane--these are all examples of linear transformations. A linear transformation[5] ($\hat{T}$) takes each vector in a vector space and "transforms" it into some other vector ($|\alpha\rangle \rightarrow |\alpha'\rangle = \hat{T}|\alpha\rangle$), with the proviso that the operation is linear: $$\hat{T}(a|\alpha\rangle + b|\beta\rangle) = a(\hat{T}|\alpha\rangle) + b(\hat{T}|\beta\rangle),\tag{3.29}$$ for any vectors $|\alpha\rangle$, $|\beta\rangle$ and any scalars $a$, $b$.

If you know what a particular linear transformation does to a set of basis vectors, you can easily figure out what it does to any vector. For suppose that $$ \begin{align} \hat{T}|e_1\rangle &= T_{11}|e_1\rangle + T_{21}|e_2\rangle + \cdots + T_{n1}|e_n\rangle \\ \hat{T}|e_2\rangle &= T_{12}|e_1\rangle + T_{22}|e_2\rangle + \cdots + T_{n2}|e_n\rangle \\ &\cdots \\ \hat{T}|e_n\rangle &= T_{1n}|e_1\rangle + T_{2n}|e_2\rangle + \cdots + T_{nn}|e_n\rangle \\ \end{align} $$ or, more compactly, $$\hat{T}|e_j\rangle = \sum_{i=1}^n T_{ij}|e_i\rangle,\quad(j = 1, 2, \ldots, n)\tag{3.30}$$

[5] In this chapter I'll use a hat ($\hat{}$) to denote linear transformations; this is not inconsistent with my earlier convention (putting hats on operators), for (as we shall see) our operators are linear transformations.


1 Answer 1


I would expect that he implies the multiplication of a linear transformation $\hat T$ with the basis vectors which are set in a column

This does not really make sense. Here, Griffiths is applying $\hat T$ to each basis vector $|e_i\rangle$ individually (on the left because that is how linear transformation act on vectors). At this point, it does not really make sense to ask whether the basis vector are arranged in a line or a column.

Doing this, we get the coefficients of $\hat T$, defined by : $$\hat T|e_j\rangle = \sum_{i=1}^n T_{ij}|e_i\rangle$$ Now, if $|x\rangle = \sum_{j=1}^n x_j |e_j\rangle$ is an arbitrary vector, then by linearity we have : $$\hat T |x\rangle = \hat T \left(\sum_{j=1}^n x_j|e_j\rangle\right) = \sum_{i=1}^n \left(\sum_{j=1}^n T_{ij}x_j\right)|x_i\rangle$$

We see that if we arrange the coefficients $T_{ij}$ in a matrix $T$ and the coordinates of $|x\rangle$ in a column vector $X$, then the coordinates of $\hat T|x\rangle$ are the coefficient of the column vector $TX$ (with the matrix multiplication by $T$ acting on the left, as expected).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.