Day: May 30, 2023

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202305301123 Exercise 5.1

The operators \hat{Q}_1, \hat{Q}_2, and \hat{Q}_3 are defined to act as follow upon a (well-behaved) function f(x):

\hat{Q}_1 squares the first derivative of f(x);
\hat{Q}_2 differentiates f(x) twice;
\hat{Q}_3 multiplies f(x) by x^4.

(a) For each of these operators, write down an explicit expression for \hat{Q}_i\,f(x).
(b) Simplify the operators expression \hat{Q}_2\hat{Q}_3-\hat{Q}_3\hat{Q}_2, writing your result in terms of the variable x. Do \hat{Q}_2 and \hat{Q}_3 commute?
(c) Derive an expression for \hat{Q}_2\hat{Q}_3-\hat{Q}_3\hat{Q}_2
in terms of the operators \hat{x} and \hat{p}.
(d) An operator \hat{Q} is linear if, for arbitrary well-behaved functions f_1(x) and f_2(x), the following property holds:

\hat{Q}\,[c_1f_1(x)+c_2f_2(x)]=c_1\,\hat{Q}\,f_1(x)+c_2\,\hat{Q}\,f_2(x).

Prove whether or not each of the three operators defined above is linear.

Extracted from M. A. Morrison. (1990). Understanding Quantum Physics A User’s Manual.

Roughwork.

(a)

\begin{aligned} \hat{Q}_1\,f(x) & =\bigg[\frac{\mathrm{d}}{\mathrm{d}x}f(x)\bigg]^2 \\ \hat{Q}_2\,f(x) & = \frac{\mathrm{d}^2}{\mathrm{d}x^2}f(x) \\ \hat{Q}_3\,f(x) & = x^4f(x) \\ \end{aligned}

(b)

If \hat{Q}_1\hat{Q}_2=\hat{Q}_2\hat{Q}_1, we say \hat{Q}_1 and \hat{Q}_2 are commuting operators to each other.

Recall addition and subtraction of operators:

\begin{aligned} (\hat{Q}_1\pm\hat{Q}_2)\,f(x) & = \hat{Q}_1\,f(x)\pm\hat{Q}_2\,f(x) \\ & = \pm\hat{Q}_2\,f(x)+\hat{Q}_1f(x) \\ \end{aligned}

so as to write

\begin{aligned} & \quad\enspace (\hat{Q}_2\hat{Q}_3-\hat{Q}_3\hat{Q}_2)\,f(x)\\ & = (\hat{Q}_2\hat{Q}_3)\,f(x)-(\hat{Q}_3\hat{Q}_2)\,f(x) \\ & = \hat{Q}_2\,(\hat{Q}_3\,f(x))-\hat{Q}_3\,(\hat{Q}_2\,f(x)) \\ & = \hat{Q}_2\,(x^4f(x)) -\hat{Q}_3\,(f''(x)) \\ & = (x^4f''(x)+12x^2f(x)) - x^4f''(x) \\ & = 12x^2f(x) \\ & \neq 0 \\ \end{aligned}

Therefore \hat{Q}_2 and \hat{Q}_3 do not commute.

(c)

The position operator, \hat{x}, in the x-representation, multiplies f(x) by x; and the momentum operator, \hat{p}_x, by \displaystyle{-i\hbar\frac{\partial}{\partial x}}:

\begin{aligned} |f\rangle & \stackrel{|x\rangle}{\longrightarrow} \langle x|f\rangle = f(x) \\ \hat{x}|f\rangle & \stackrel{|x\rangle}{\longrightarrow} \langle x|\hat{x}|f\rangle = xf(x) \\ \hat{p}|f\rangle & \stackrel{|x\rangle}{\longrightarrow} \langle x|\hat{p}|f\rangle = -i\hbar\nabla f\\ \end{aligned}

\hat{Q}_2\hat{Q}_3-\hat{Q}_3\hat{Q}_2=\textrm{I don't know?}

Note the symmetry between the x and the p representations and a one-to-one mapping onto each other.

(d) Not to be attempted.