-4
$\begingroup$

In general when we study the expectation value of any variable, which is a function of position and momentum in the way$$Q(x,p)$$ We generally do the following thing,i.e. $$\int_{x=0}^\infty \psi^*Q(x,p)\psi dx$$

If p be its momentum,and we are to determine $p^2$,we know,momentum operator is defined as $-ih\frac{d}{dx}$ So to determine the expectation value of p^2 we will simply replace Q in the above way,where we assume $(\frac{d}{dx})^2=\frac{d^2}{dx^2}$. Now my question is how can we understand that replacing the differential operators in such a way indeed begets us the expectation values of the required? i.e. $$<p^2>= \int_{x=0}^\infty \psi^* (-h^2\frac{d^2}{dx^2})\psi dx$$

$\endgroup$
  • 1
    $\begingroup$ $<Q^2> \neq <Q>^2$. Trivial example: $\{1,-1\}$ has an expectation value of $0$ while the expectation value of their squares is $1$. $\endgroup$ – user121330 Oct 12 '17 at 17:04
  • 3
    $\begingroup$ I don't understand what you are asking! "we will simply replace Q in the above way,where we assume... " - That is the definition of the operator. We do not assume that $(\frac{d}{dx})^2 = \frac{d^2}{dx^2}$. It is how $p^2$ is defined. First act $p$ on $\psi$ then act $p$ again. Each action of $p$ is a derivative. I am unclear on what you are asking $\endgroup$ – Prahar Oct 12 '17 at 18:57
1
$\begingroup$

An expectation value $<Q>$ of an observable $Q$ in quantum mechanics is always defined with respect to a state $\psi$ in the manner $<Q> = <\psi|Q|\psi>$.

All observables are Hermitian and have a complete set of eigenstates. Let us denote these by $|q_i>$. Because the set $|q_i>$ is complete, the identity operator may be written as $1 = \sum_i |q_i><q_i|$. Hence

$$ <Q>_\psi \equiv <\psi|Q|\psi> = \sum_{i,j} <\psi|q_j><q_j|Q|q_i><q_i|\psi> $$

$$ =\sum_{i,j} Q_i\delta_{i,j}<\psi|q_j><q_i|\psi> = \sum_iQ_i<\psi|q_i><q_i|\psi> $$

$$ =\sum_iQ_i<q_i|\psi><q_i|\psi>^* = \sum_iQ_i|<q_i|\psi>|^2. $$ Because $|<q_i|\psi>|^2$ is the probability of finding the system in state $|q_i>$ given that it is in state $|\psi>$, it is evident that the last expression is the usual definition of the expectation value of the measurement of the observable $Q$ on the state $|\psi>$.

When we are dealing with a continuum of states rather than a discrete set, as is the set of position states $|x>$, the identity operator is written now as $\int dx|x><x|$. Hence

$$ <Q>_\psi \equiv <\psi|Q|\psi> = \int dx_1\int dx_2<\psi|x_2><x_2|Q|x_1><x_1|\psi> $$

$$ \equiv \int dx_1\int dx_2 \psi^*(x_2)<x_2|Q|x_1>\psi(x_1). $$

If $Q$ is a function of position $x$,

$$ <x_2|Q(x)|x_1> = <x_2|Q(x_1)|x_1> = Q(x_1)<x_2|x_1> = Q(x_1)\delta(x_2 - x_1) $$

and it follows that

$$ <Q>_\psi = \int dx_1\psi^*(x_1)Q(x_1)\psi(x_1). $$ If $Q$ is a function of momentum, we must use the position-space representation of $p$, which is $p = -i\hbar\partial/\partial x$. This is a result from the fact that momentum is the generator of translations in position space. For a full derivation of said result, I recommend Townsend's book "A Modern Approach to Quantum Mechanics", chapter 6. If I recall correctly, mainly it follows from the result that the position and momentum representation form a Fourier transform pair, i.e.

$$ <x|p> = \frac{1}{\sqrt{2\pi\hbar}}\exp(ipx/\hbar). $$

I'll come back later with the proof if necessary.

$\endgroup$
1
$\begingroup$

Clearly $\langle A\rangle^2\ne \langle A^2\rangle$. In your specific case, $p$ can take positive and negative values for $p^2$ is non-negative.

As an analogy, consider a 6-sided die. The average throw is $$ \frac{1}{6}(1+2+3+4+5+6)=\frac{7}{2} $$ and the square of this is $49/4=12.25$ but the average of the squares $$ \frac{1}{6}(1+2^2+3^2+4^2+5^2+6^2) \approx 230. $$

Similarly, imagine a 2-sided die with values $-1$ and $1$. The average would be 0, so the square of that is also $0$, but the average of the squares would be $1$.

$\endgroup$
  • $\begingroup$ Actually I was not looking for that,edited my question now.I tried to ask for something different,may be I failed to clarify. $\endgroup$ – user157588 Oct 12 '17 at 17:24
  • $\begingroup$ @user157588 Please just make sure you're clarifying your question, and not changing it to ask something different. If you're editing in a way that makes existing answers invalid, that's usually a sign that your edit may have gone too far. $\endgroup$ – David Z Oct 12 '17 at 18:32

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.