Welcome to our community

Be a part of something great, join today!

Probability and Operators in Quantum Mechanics

  • Thread starter
  • Admin
  • #1


Indicium Physicus
Staff member
Jan 26, 2012
Unfortunately, I can't find the thread (if someone finds it, please let me know, and I'll merge this post onto that thread), but someone asked why it is that in quantum mechanics, if you have an observable $B$, that the expectation value (average value) $\langle B \rangle$ is given by
$$\langle B \rangle = \int_{\mathbb{R}}\Psi^{*} \hat{B} \, \Psi\,dx,$$
where $\hat{B}$ is the operator "associated with" the observable $B$.
If I recall, the observable in question was the momentum operator
$$\hat{p}= -i \hbar\,\frac{\partial}{ \partial x}.$$
So, why is
$$ \langle p \rangle= \int_{\mathbb{R}}\Psi^{*} \left( -i \hbar\,\frac{\partial}{ \partial x} \right)\, \Psi\,dx?$$

The following derivation will follow Griffiths' Introduction to Quantum Mechanics, 1st Ed., pages 11-16.

Let's start with $\Psi$, which is the wave function solution of the time-dependent Schrödinger equation in one dimension:
$$i \hbar \frac{ \partial \Psi}{ \partial t}=- \frac{ \hbar^{2}}{2m}
\frac{ \partial^{2} \Psi}{\partial x^{2}}+V \Psi.$$
The statistical interpretation of the Schrödinger equation tells us that the wave function $|\Psi(x,t)|^{2}$ is the probability density for finding the particle at point $x$ at time $t$. So if I want to find the expectation value of the observable $x$, I should do
$$\langle x \rangle=\int_{\mathbb{R}}x|\Psi(x,t)|^{2}\,dx.$$
Now, if we want the observable $p$, we would like
Shifting to expectation values, we would like
$$\langle p \rangle=m\frac{d \langle x \rangle}{dt}
=m\int_{\mathbb{R}}x \frac{ \partial}{ \partial t}| \Psi(x,t)|^{2}\,dx.$$
Now the Schrödinger equation tells us that
$$ \frac{ \partial \Psi}{\partial t}= \frac{i \hbar}{2m} \frac{ \partial^{2} \Psi}{ \partial x^{2}}- \frac{i}{ \hbar} V \Psi,$$
and hence
$$ \frac{ \partial \Psi^{*}}{\partial t}=- \frac{i \hbar}{2m} \frac{ \partial^{2} \Psi^{*}}{ \partial x^{2}}+ \frac{i}{ \hbar} V \Psi^{*}.$$
$$\frac{ \partial}{ \partial t}| \Psi(x,t)|^{2}
=\frac{ \partial}{ \partial t}(\Psi^{*} \Psi)
= \Psi^{*} \frac{ \partial \Psi}{ \partial t}+ \Psi \frac{ \partial \Psi^{*}}{ \partial t}$$
$$=\Psi^{*} \left(\frac{i \hbar}{2m} \frac{ \partial^{2} \Psi}{ \partial x^{2}}- \frac{i}{ \hbar} V \Psi \right)+
\Psi \left( - \frac{i \hbar}{2m} \frac{ \partial^{2} \Psi^{*}}{ \partial x^{2}}+ \frac{i}{ \hbar} V \Psi^{*} \right)$$
$$= \frac{i \hbar}{2m} \left( \Psi^{*} \frac{ \partial^{2} \Psi}{ \partial x^{2}}- \Psi \frac{ \partial^{2} \Psi^{*}}{ \partial x^{2}} \right)$$
$$= \frac{i \hbar}{2m} \frac{ \partial}{ \partial x}\left( \Psi^{*} \frac{ \partial \Psi}{ \partial x}- \Psi \frac{ \partial \Psi^{*}}{ \partial x} \right).$$
Plugging this into our latest expression for $\langle p \rangle$ yields
$$\langle p \rangle=m\int_{\mathbb{R}}x \frac{ \partial}{ \partial t}| \Psi(x,t)|^{2}\,dx
=m\int_{\mathbb{R}}x \left( \frac{i \hbar}{2m} \frac{ \partial}{ \partial x}\left( \Psi^{*} \frac{ \partial \Psi}{ \partial x}- \Psi \frac{ \partial \Psi^{*}}{ \partial x} \right) \right)\,dx$$
$$=\frac{i \hbar}{2}\int_{\mathbb{R}}x \frac{ \partial}{ \partial x}\left( \Psi^{*} \frac{ \partial \Psi}{ \partial x}- \Psi \frac{ \partial \Psi^{*}}{ \partial x} \right) \,dx.$$
We can integrate this by-parts:
$$ \langle p \rangle=-\frac{i \hbar}{2}\int_{\mathbb{R}} \left( \Psi^{*} \frac{ \partial \Psi}{ \partial x}- \Psi \frac{ \partial \Psi^{*}}{ \partial x} \right) \,dx.$$
Here we have used the fact that $\Psi$ must go to zero as $x\to \pm \infty$ to eliminate the boundary term. Now, if we integrate the second term by-parts, we will obtain
$$ \langle p \rangle =-i \hbar \int_{ \mathbb{R}} \Psi^{*} \frac{ \partial \Psi}{ \partial x} \, dx
=\int_{\mathbb{R}}\Psi^{*} \left( -i \hbar\,\frac{\partial}{ \partial x} \right)\, \Psi\,dx.$$
As it turns out, all observables can be written in terms of position and momentum, so everywhere you see an $x$, "replace" it with multiplication by $x$, and everywhere you see a $p$, replace it with the operator $-i \hbar (\partial / \partial x)$.

So, to sum up: this is the operator representation of $p$, because we want to impose the condition that $\langle p \rangle= m d \langle x \rangle/dt$. And then, because of the Schrödinger equation and the derivation I showed above, you obtain the desired representation.