1
$\begingroup$

We use in Dirac formalism of QM the tool of operators and kets in spatial and spin spaces to obtain eigenvalues and eigenkets.

But the operation here is simply that of a matrix multiplication.

Now the question is that why is the emphasis on saying an "operator" instead of simply a matrix.

Are there examples in which we cannot have a matrix representation?

Or it is because of some abstract algebra involving group theory and "operations" or mapping on vectors?

Put the question another way, for someone unfamiliar with group theory, is it reasonable to replace the object of an operator with a matrix in a textbook without losing much in Dirac formalism of QM theory?

$\endgroup$
5
$\begingroup$

Now the question is that why is the emphasis on saying an "operator" instead of simply a matrix.

A matrix is a representation of an operator expressed in a particular basis. Consider the operation $T$ which mirrors the 2D plane about the line $x=y$. If we construct the obvious basis vectors $\hat{x}$ and $\hat{y}$, then $T$ is represented as $$[T]_{xy} = \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right]$$ where $[\cdot]_{xy}$ means "representation in the $\hat{x}\hat{y}$ basis. To see that this is true, consider the action of $T$ on $\hat{x}$. Flipping $\hat{x}$ about the line $y=x$ gives $\hat{y}$. Since $\hat{x}$ is represented as $$[x]_{xy} = \left[ \begin{array}{c} 1 \\ 0 \end{array} \right]$$ we get $$[T]_{xy} [x]_{xy} = \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right] \left[ \begin{array}{c} 1 \\ 0 \end{array}\right] = \left[ \begin{array}{c} 0 \\ 1 \end{array}\right] = [y]_{xy} \, . $$ You can check that the case for $T$ acting on $\hat{y}$ works too. Now suppose we use a different basis. Define $$ \hat{p} \equiv \frac{1}{\sqrt{2}} \left( \hat{x} + \hat{y} \right) \quad \text{and} \quad \hat{q} \equiv \frac{1}{\sqrt{2}} \left( \hat{x} - \hat{y} \right) \, . $$ Now you can pretty clearly envision that $T\hat{p} = \hat{p}$ and $T\hat{q} = - \hat{q}$. Therefore, the matrix for $T$ in the $\hat{p}\hat{q}$ basis is $$[T]_{pq} = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right] \, .$$ This is a different matrix from the one we had in the $\hat{x}\hat{y}$ basis.$^{[a]}$

The emphasis on "operator" is supposed to remind you that the thing you're dealing with is more general than any one possible matrix representation. Keeping this in mind and using notation which brings that out is extremely helpful in understanding how to solve lots of problems.

Are there examples in which we cannot have a matrix representation?

Yes. A very simple example is the derivative. First, note that functions are vectors; they satisfy all properties of a vector space:

  • Adding two functions gives another function.

  • Multiplication of a function by a scalar gives another function.

  • There is a zero function, which when added to another function $r$ yields $f$.

Since functions are vectors you might expect that we can construct linear operators on the set of functions. The derivative is a linear operator because $D_x(f + g) = D_x f + D_x g$ where here $D_x$ means the derivative with respect to $x$. Thus $D_x$ is a linear operator. However, it's not always possible to write it as a matrix, and it's never possible to write it as a finite dimensional matrix.

Put the question another way, for someone unfamiliar with group theory, is it reasonable to replace the object of an operator with a matrix in a textbook without losing much in Dirac formalism of QM theory?

No! It's important to keep the Dirac notation. Any matrix representation is just one possible representation. It's often much easier to solve problems without going to a particular basis until the end of the calculation.

$[a]$: Note that $[T]_{pq}$ is diagonal. I did that on purpose. If you pick your basis vectors right you can often find a basis in which the matrix representation of an operator is diagonal. A lot of physics problems ranging from coupled oscillators to expansions of electric potential, to solutions to Schrodinger's equation rely heavily on diagonalizing an operator. The physical states represented by the vectors which make up the diagonalizing basis are called "normal modes".

$\endgroup$
2
$\begingroup$

The same thing happens in 3d you can have an operator of rotation of a quarter turn about a particular vector.

Or you can have a matrix.

But the latter requires a choice of basis. The former does not. In quantum mechanics, you often want to pick the basis after you know the operator and if you wrote every operator as a diagonal matrix with real numbers down the diagonal, the whole lack of commutativity would be hard to describe.

Plus. If you are working with an infinite dimensional space instead of a finite dimensional space then using new language can remind you not to just carry over every result and to only carry over the things that still hold.

$\endgroup$

Your Answer

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