Extra Credit 47

Expectation, Momentum, and Uncertainty

Our job now is to properly define the uncertainties Δx and Δp.

As an aside, let us review the properties of discrete probability distributions.

Consider the number distribution N of ages a in a population. The probability of finding a person with a given age is \(\mathbb{P}(a) = \frac{N(a)}{N_{total}}\) satisfying \(\sum _{a}\mathbb{P}(a)=1\)

What is the most likely age? In this case, that is 22.

What is the average age? In general, the weighted average

\(\left \langle a \right \rangle=\frac{\sum_{a} aN(a)}{N_{total}} = \sum_{a} a\mathbb{P}(a).\)

In this case, it is 19.4. Note that in general, as in this example, (a) does not have to be a measurable value of a!

What is the average of the squared age? In general,

\(\left \langle a^{2} \right \rangle =\sum_{a} a^{2}\mathbb{P}(a).\)

For a general function of the age,

\(\left \langle f(a) \right \rangle = \sum_{a} f(a)\mathbb{P}(a).\)

Is the average of the squared age equal to the square of the average age? In mathematical notation, is \(\left \langle a^{2} \right \rangle = \left \langle a \right \rangle^{2}\)? No! If a represented a more general quantity rather than age, it could sometimes be positive or negative, and those terms might cancel out in the average. By contrast, a² would never be negative, so its average would satisfy that too.

How do we characterize the uncertainty? We could use \(\Delta a= a -\left \langle a \right \rangle\), but the problem is that \(\left \langle\Delta a \right \rangle = 0\) identically. Instead, we use the standard deviation defined by

\(\left \langle\Delta a \right \rangle^{2} =\left \langle (a-\left \langle a \right \rangle)^{2} \right \rangle\),

which also satisfies

\(\left \langle\Delta a \right \rangle^{2} = \left \langle a^{2} \right \rangle - \left \langle a \right \rangle^{2}.\)

In this case, the standard deviation is about 2.8.

Similar expressions exist for continuous variables. Given that ψ has been discussed as a function of position x thus far, it makes sense to proceed in that way. Mathematically,\[\left \langle\ f(x) \right \rangle =\int_{-\infty }^{\infty } f(x)\mathfrak{p} (x) dx \] but \(\mathfrak{p} (x) = \psi ^{*}(x)\psi (x).\) Hence, the way to find the expectation value of a function of position in a given quantum state is \[\left \langle\ f(x) \right \rangle =\int_{-\infty }^{\infty } \psi ^{*}(x)f(x)\psi (x) dx.\]

In all this, the normalization \(\int_{-\infty }^{\infty } \mathfrak{p} (x)dx= 1\) is assumed. From this, the uncertainty in position \[\Delta x\equiv \sqrt{\left \langle x^{2} \right \rangle- \left \langle x \right \rangle^{2}}\] can be found.

Notice that expectation values \(\left \langle f(x) \right \rangle\) depend on the state! This can be written as \(\left \langle f(x) \right \rangle_{\psi } , \left \langle f(x) \right \rangle_{|\psi } , or \left \langle \psi | f(x)|\psi \right \rangle\)

For example, let us consider a wavefunction given by \[\psi (x)= \left \{ N\cdot \left ( ^{2} -l^{2}\right )^{2} for |x| \leq l, 0 otherwise \right \} \]

We need to figure out the normalization for this wavefunction by \[ \int_{-\infty }^{\infty } |\psi (x)|^{2}dx = 1 \]

which, when effected by nondimensionalization of the integral, yields \(N = \sqrt{\frac{315}{256}}\frac{e^{i\varphi }}{\sqrt{l} }.\)

After this, by noting that \(|\psi (x)|^{2}\) is even while x is odd , then \(\left \langle x \right \rangle = 0\). Also, \(\left \langle x^{2} \right \rangle = \frac{l^{2}}{11}\). Hence, \(\Delta x = \frac{t}{\sqrt{11}}.\)

After all of this, how do we find the momentum expectation value \(\left \langle p \right \rangle\) Naively, we might say that \(\left \langle p \right \rangle = \int_{-\infty }^{\infty }\psi ^{*}(x) p\psi (x)dx.\) But how exactly are we to express p in an integral over functions of x? Clearly, this will not do!

Here’s a hint: we know that a wave with

\(k=2\pi \lambda ^{-1} \)

is associated with a particle with

\(p = h \lambda ^{-1} = \hslash\ k .\)

Disregarding normalization, the associated wavefunction is

\(\psi = e^{ikx}.\)

But note that

\(\frac{\partial e^{ikx} }{\partial x} = ike^{ikx}. \)

This means that

\(-i\hslash \frac{\partial e^{ikx} }{\partial x} = \hslash ke^{ikx}. \)

Thus

\(-i\hslash \frac{\partial e^{ikx} }{\partial x} = p \cdot e^{ikx},\)

and the units work out too! But what does momentum have to do with a derivative with respect to position anyway?

Here’s another hint: Noether’s theorem states that to every symmetry is associated a conserved quantity.

So momentum is associated with spatial translations! Now consider how translations behave for functions:

\[f(x)\rightarrow f(x+l)=f(x) + l\frac{\partial f(x)}{\partial x} + \frac{l^{2}}{2}\frac{\partial^2 f(x)}{\partial x^2} + ... \]

\[=\sum_{n=0}^{\infty } \left ( l\frac{\partial }{\partial x} \right )^{n} f(x)\]

\[= e^{\frac{\partial l}{\partial x}} f(x) \]

Hence translations are generated by spatial derivatives \(\frac{\partial }{\partial x}\) . But we just said that translations are associated with p! This means that it is natural to associate p with \(\frac{\partial }{\partial x}\) somehow. In a ∂x similar way, E would be associated with \(\frac{\partial }{\partial t} \), and L_z with \(\frac{\partial }{\partial \varphi } \).

That’s enough for hints. We need to take a stand on this.

Momentum in quantum mechanics is realized by an operator

\[\hat{p} = -i \hslash \frac{\partial }{\partial x} . \]

This operator \(\hat{p} \) is what we use to compute expectation values. More precisely, \[\left \langle p^{2} \right \rangle = \left ( -i \hslash \right )^{n} \int_{-\infty }^{\infty } \psi ^{*} (x) \frac{\partial^n \psi (n)}{\partial x^n} dx \]

and the uncertainty is then given by

\(\Delta p = \sqrt{\left \langle p^{2} \right \rangle - \left \langle p \right \rangle^{2}} \)

Let us return to our previous example wavefunction given by

\[\psi (x) = \left \{ N\cdot \left ( x^{2} - l^{2} \right )^{2} for |x| \leq l, 0 otherwise \right \}.\]

Now we can find \[\left \langle p \right \rangle = -i\hslash \int_{-\infty }^{\infty } \psi^{*} (x) \frac{\partial \psi (x)}{\partial x} dx\]

\[= -i \hslash |N|^{2}\int_{-\infty }^{\infty } \left ( x^{2}-l^{2} \right )^{2} \cdot (2\cdot 2x\cdot \left ( x^{2}- l^2 \right ))dx \]

\[=0\]

as the wavefunction is even while its spatial derivative is odd.

By a similar computation, \(\left \langle p^{2} \right \rangle = \frac{3\hslash^{2}}{l^{2}} \), which dimensionally makes sense as well. From this, we find that \(\Delta p = \frac{\sqrt{3} \hslash}{l} \), and the uncertainty relation is satisfied as \( \Delta x\Delta p = \sqrt{\frac{3}{11}} \hslash. \).

But what does this new operator \(\hat{p}\) have to do with having momentum \(p = \hslash k?\) Let us consider two states given by

\(\psi _{k}(x)= e^{ikx}\)

\(\psi _{s}(x)= e^{ikx} + e^{ik{'}x}. \)

The first has definite momentum \(p = \hslash k\) , while the second, being a superposition of states with definite momenta \(p = \hslash k\) and \(p^{'} = \hslash k^{'}\), is not itself a state of definite momentum. We can show this by acting on each state with the operator \(\hat{p}\):

\(\hat{p}\psi_{k} (x) = \hslash k e^{ikx}\)

is simply proportional to \(\psi _{k}(x)\), while

\(\hat{p}\psi _{s}(x)= \hslash \cdot (ke^{ikx} + k^{'}e^{ik'x})\)

is not simply proportional to \(\psi _{s}(x)\). We see that \(\hat{p}\) is an operator which acts simply on wavefunctions corresponding to states with definite momenta, but not on arbitrary superpositions of momentum states. This means that \(\hat{p}\) is the operator whose eigenstates are states of definite momentum, and the corresponding eigenvalue is exactly the momentum of that state.