Question

1.30 The translation operator for a finite (spatial) displacement is given by where p is the momentum operator. (a) Evaluate (b) Using (a) (or otherwise) demonstrate how the expectation value (x) changes under translation.

please explain in detail.

Answer 1

Let us start by writing down some important stuff.

$\hat{p_x} = -i\hbar \frac{\mathrm{d} }{\mathrm{d} x}$

$\hat{\vec{p}} = -i\hbar \frac{\mathrm{d} }{\mathrm{d} \vec{r}}$

Imagine have a wavefunction. This wavefunction describes a system. Doesn't matter what shape it is. What matters is its position. Now, we move the system (the entire system) without changing the shape or any property of the function to the right, by a distance . How do we describe this using the rules of quantum mechanics?

Well, any change or modification or measurement, etc in QM is done using operators. For this too, we need an operator. Let us define one. But first,

$\Psi(x) = initial \ state=|x\rangle$

$\Psi(x+l) = \Psi'(x)=final \ state=|x'\rangle = |x+l\rangle$

Therefore

$\Psi'(x) = \Psi(x-l) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

Now finally defining our translation operator, such that the final state is obtained after we operate on the initial state.

$| x' \rangle = g(l)|x\rangle$

this g(l) is our translational operator, which moves our function to the right by a distance . To find it's value, let's expand the RHS of equation ,

$\Psi'(x) = \Psi(x) - l.\frac{\mathrm{d} }{\mathrm{d} x} \Psi(x)+ \frac{l^2}{2!}\frac{\mathrm{d}^2 }{\mathrm{d} x^2}\Psi(x)+....$

$\Psi'(x) = \left \{ \1 - l.\frac{\mathrm{d} }{\mathrm{d} x}+ \frac{l^2}{2!}\frac{\mathrm{d}^2 }{\mathrm{d} x^2}+.... \right \}\Psi(x)$

We know that the series inside the brackets is an exponential series, so, we can write,

$\Psi'(x) = exp \left \{ -l\frac{\mathrm{d} }{\mathrm{d} x} \right \}\Psi(x)$

Now we know that , $\hat{p_x} = -i\hbar \frac{\mathrm{d} }{\mathrm{d} x}$ ,

There we can write, in bra-ket notation,

$|x'\rangle = exp\left \{ -i\frac{l}{\hbar}\hat{p}_x \right \}|x\rangle$

This is perfect. We have our operator.

$g(l) = exp\left \{ -i\frac{l}{\hbar}\hat{p}_x\right \}$

If we generalize to the 3-D case, we can easily make all the quantities to be vectors, giving us,

$g(\vec{l}) = exp\left \{ -\frac{i}{\hbar}\vec{l}.\vec{p} \right \}$

Similarly we can also derive the hermitian conjugate of our operator,

$[g(\vec{l})]^{\dagger} = exp\left \{ \frac{i}{\hbar}\vec{l}.\vec{p} \right \}$

$|x\rangle = exp\left \{ i\frac{l}{\hbar}\hat{p}_x \right \}|x'\rangle$

This moves the system to the opposite side by a distance .

(A) Now, let us write the commutator relation we are asked to derive. Let's derive it for a single dimension and then generalize it for higher dimensions, since the position and momentum of different axis commute.

$\left [ \hat{x},g(l) \right ] = g(l). \hat{x} - \hat{x}.g(l)$

$g(l). \hat{x}|x\rangle = xg(l) |x\rangle = x|x'\rangle$

$\hat{x}.g(l)|x\rangle = \hat{x}|x'\rangle = (x+l)|x+l\rangle$

Using these values, we can write,

$\left [ \hat{x},g(l) \right ]|x\rangle = (x+l)|x+l\rangle - x|x'\rangle$

             $= (x+l)|x'\rangle - x|x'\rangle$

                 $= (x+l)g(l)|x\rangle - xg(l)|x\rangle$

                 $= ((x+l)g(l)- xg(l))|x\rangle$

                 $= l.g(l)|x\rangle$

$\left [ \hat{x},g(l) \right ] = l.g(l)$

Generalizing in 3-D we get,

$\left [ \vec{l},g(\vec{l}) \right ] = l_i.g(\vec{l})\ where \ i = x,y,z$

(B) To see how the expectation value (mean of all of the observed value) of position is affected, we write,

$\langle \hat{x}\rangle' = \langle \Psi'|\hat{x}|\Psi'\rangle = \langle \Psi|[g(l)]^{\dagger}\hat{x}.g(l)|\Psi\rangle$

                                   $= \langle \Psi|[g(l)]^{\dagger}\left [ \hat{x},g(l) \right ]|\Psi\rangle +\langle\Psi|[g(l)]^{\dagger}g(l). \hat{x}|\Psi\rangle$

                                   $= \langle \Psi|[g(l)]^{\dagger}l.g(l)|\Psi\rangle +\langle\Psi|[g(l)]^{\dagger}g(l). \hat{x}|\Psi\rangle$

                                    $= l\langle \Psi|\Psi\rangle +\langle\Psi|\hat{x}|\Psi\rangle\ \ \ \ \ \ \ \ \ \ \because g(l)[g(l)]^\dagger= [g(l)]^{\dagger}g(l)=1$

                                   $= \langle \Psi|\hat{x}|\Psi\rangle + l$

                                 $= \langle \hat{x}\rangle + l$

As you can see, the expectation value of the position increase in magnitude by .