Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm having trouble understanding the motivation for the definition of the expectation of a self adjoint operator $A$:

$$\langle A \rangle _\psi=\int_{\mathbb{R}}\psi^*A\hspace{0.2cm} \psi \hspace{0.2cm} dx$$

where $\psi(x,t)$ is a normalised state.

I can understand the expectation of the position operator in terms of basic probability: one of the assumptions of QM being that $|\psi|^2$ is the probability to find the particle at $x$. The expected value is just the sum of the positions multiplied by their probabilities:

$$\langle x \rangle _\psi=\int_{\mathbb{R}}x \hspace{0.1cm} |\psi(x,t)|^2 dx=\int_{\mathbb{R}}\psi^* \hspace{0.2cm} x \hspace{0.1cm} \psi \hspace{0.1cm}dx$$ by commutativity.

I don't understand how this might work with the momentum operator, for example. The lecture notes for my course say that the expectation of self adjoint operators are defined in analogy to this, so that

$$\langle p \rangle _\psi=\int_{\mathbb{R}}\psi^*\hspace{0.1cm}p \hspace{0.1cm}\psi\hspace{0.1cm} dx$$

I can't satisfy myself with this explanation; $p$ is a differential operator and so can't be moved about within the integral like $x$ can. For example

$$\int_{\mathbb{R}}\psi^*\hspace{0.1cm}\psi\hspace{0.1cm} p\hspace{0.1cm} dx$$ makes no sense to me, as the operator hasn't been applied to anything. In any case, $\int_{\mathbb{R}}|\psi(x,t)|^2 p dx$ doesn't mean anything to me probability-wise.

So basically I'm looking for an explanation as to why the expectation of self-adjoint operators are so defined. Thanks for any replies!

share|improve this question

2 Answers 2

up vote 0 down vote accepted

It is very common to abuse the notation here, so I'll try to clarify a bit. The state of a physical system can be described by an abstract vector $\left|\psi\right\rangle$, which is an element of a Hilbert space. The wavefunction, $\psi(x)$ is the representation of that vector in the position basis, $\psi(x)\equiv\left\langle x | \psi\right \rangle \equiv \left\langle x , \psi\right \rangle$. In this notation, $\left|x\right\rangle$ is a state which is located at the point $x$ and nowhere else.

Think of $\left|\psi\right\rangle$ as a column vector whose components are the values of $\psi(x)$ at each $x$.

$$\psi(x) = \begin{pmatrix}\vdots\\\psi(x_1)\\\psi(x_2)\\\psi(x_3)\\\vdots\end{pmatrix}$$

This vector also has a dual, $\left\langle\psi\right| = \left|\psi\right\rangle^\dagger$, which is the transpose conjugate.

In a different basis, this vector would have different components. Another common basis is the momentum basis, with components $\psi_p(p)=\left\langle p | \psi\right\rangle$, typically written as simply $\psi(p)$.

Just like we could write a 3d vector $\bf{v}$ in the $\bf{i},\bf{j},\bf{k}$ basis as ${\bf v} = {\bf v}_1{\bf i} + {\bf v}_2{\bf j} + {\bf v}_3{\bf k}$, we can write $\left|\psi\right\rangle$ as a sum of $x$ components,

$$\begin{align} \left|\psi\right\rangle&= \int^{\infty}_{-\infty}\left|x\right\rangle\,\psi(x)\,\text{d}x \\ &= \int^{\infty}_{-\infty}\left|x\right\rangle\!\left\langle x | \psi\right \rangle \,\text{d}x \end{align}$$

From this it is clear that $\int^{\infty}_{-\infty}\left|x\right\rangle\!\left\langle x \right|\text{d}x$ is an identity, since when it acts on $|\psi\rangle$ it gives $|\psi\rangle$ back.

Eigenvectors are usually labeled by their eigenvalues, so that if $\hat p$ is an operator with an eigenvalue $p$, we write $\hat{p}\left|p\right\rangle = p\left|p\right\rangle$. The eigenvectors of a self-adjoint operator form a complete set of states, which is to say that the sum of all the projection operators is the identity, $\int \left|p\right\rangle\left\langle p\right| \,\text{d}p $. This means you could also write

$$\begin{align} \langle p\rangle&= \left\langle\psi\right|\hat{p}\left|\psi\right\rangle \\ &=\left\langle\psi\right|\hat{p}\int\left|p\right\rangle\!\left\langle p\right|\,\text{d}p\,\left| \psi\right\rangle\\ &=\int{p}\,\langle\psi\left|p\right\rangle\!\left\langle p\right| \psi\rangle\,\text{d}p\\ &=\int p\,\psi^*\!(p)\,\psi(p)\text{d}p \end{align}$$

where $\psi(p)=\left\langle p | \psi\right\rangle$. This might make more sense, since it is more obviously an average value of p.

The reason that $\hat p = -i\hbar{\partial \over \partial x}$ in the x basis is that a momentum eigenstate (a state with a single wavelength) is a plane wave, $\langle x \left|p\right\rangle = {1 \over \sqrt{2\pi}} e^{ipx/\hbar}$, and that means that in the x basis, $\left\langle x \right| \hat p\left|p\right\rangle = {p \over \sqrt{2\pi}} e^{ipx/\hbar} = {-i\hbar \over \sqrt{2\pi}}{\partial \over \partial x} e^{ipx/\hbar}$.

share|improve this answer
    
I thought that $\psi$ was a function of $x$, not $p$? Also, I can understand that $\langle p||p\rangle$ is the inner product of the state $p$ with itself, but I'm not sure what $|p\rangle \langle p|$ is. –  James Machin Mar 31 at 20:43
    
$\psi(x)$ is shorthand for $\left\langle x | \psi \right \rangle$. The state vector $\left | \psi\right\rangle$ lives in a Hilbert space, and describes the properties of the state. $|x\rangle$ and $|p\rangle$ are states with definite position and momentum, respectively. $|p\rangle\langle p|$ is a projection operator that can act on $|\psi\rangle$ to give the part of $|\psi\rangle$ with momentum p. –  George G Mar 31 at 20:50
    
I had no idea that $\psi (x)$ was shorthand for $\langle x| \psi \rangle$. I'm having trouble making sense of that, as as far as I knew, $\langle x| \psi \rangle = \langle x, \psi (x) \rangle = \int_{\mathbb{R}}x\psi (x).$ I'm not getting the distinction that you're making between $\psi (x)$ and $|\psi \rangle$. I thought that the Hilbert space was a space of functions? –  James Machin Mar 31 at 21:15
    
Also, I don't understand why $|p \rangle \langle p|$ is an operator at all. Is there a definition of $|\phi \rangle \langle \psi |$? I should also add thanks for all the help so far! –  James Machin Mar 31 at 21:16
    
There is a lot of abuse of notation going on here, which does make this confusing. I'll edit my answer with more info. –  George G Mar 31 at 22:13

You need to apply the operator first and then evaluate the integral:

$⟨P⟩_ψ = i\hbar\int{\psi^*(x)\frac{d\psi(x)}{dx}dx}$

share|improve this answer
    
Yes I know that this is the correct thing to do, but I want to know why. Why is the expectation of the momentum defined like that? –  James Machin Mar 31 at 20:31
    
I'm not sure I understand what you are asking. Its just a fancy form of a weighted average. It applies to any operator. –  PhotonicBoom Mar 31 at 20:38
    
I'm trying to understand why the expression that you've written above is a weighted average. I can't see how it is an average in any way. Maybe my knowledge of statistics is lacking? –  James Machin Mar 31 at 20:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.