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I want to understand precisely where the formula for the expectation value of $x$ comes from (in QM): $$\langle x\rangle=\int _{-\infty}^{\infty}\psi ^*x\psi dx $$

I know that an expectation value (in statistics) is just the sum of the products of the possible values $f(x)$ times their probabilities $\rho (x)$: $$\langle f(x)\rangle=\int f(x) \rho (x)dx $$ Since in QM mechanics the probability is given by $|\psi|^2 $, the expectation value of $f(x)$ would be: $$\langle f(x)\rangle=\int f(x) |\psi|^2dx=\int f(x)\psi^*\psi dx$$ But this differs from the form above. If $f(x)$ was Hermitian I could use the property of Hermitian operators to "move it" into the position that it should be, but since it is not necessarily Hermitian, I don't know how to explain this difference, or how to solve it. I have consulted Griffith's QM and also online, but I cannot find an answer.

What am I missing here?

Qmechanic
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Nick Heumann
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  • I'm a bit confused about what your question is actually about. Where do you want to "move" $f(x)$ to? – noah Jan 18 '21 at 17:04
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    $x$ in all RHSs is just an integration variable – fqq Jan 18 '21 at 17:12
  • I want the f(x) to be between the "psi"s. Because while I could move the x, if the operator were, for example, the momentum operator, then I don't think that I could just move it, since it would be acting on the product of the "psi"s – Nick Heumann Jan 18 '21 at 17:19
  • @NickHeumann maybe instead your question should be why $\langle \hat{A}\rangle = \int dx ; \psi^* \hat{A} \psi$ in general then? – Triatticus Jan 18 '21 at 17:35
  • The momentum operator is not $-i\partial_x$ is I think the confusion? That's a position space representation. – jacob1729 Jan 18 '21 at 17:36
  • Related: https://physics.stackexchange.com/questions/299286/why-does-the-expectation-value-of-an-operator-a-take-the-form-langle-a-ran?rq=1 – Tobias Fünke Jan 18 '21 at 17:44
  • Check also this: "general expectation value in QM" https://physics.stackexchange.com/a/757441/226902 – Quillo Apr 01 '23 at 13:00

3 Answers3

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Inside the integrals, everything is a scalar, you can rearrange terms as you wish. It's a bit hard to see because you omitted the $x$-dependence of $\psi$. It really is $$\int \psi^*(x)\,x\,\psi(x) dx$$ where $\psi(x) = \langle x|\psi\rangle$, which clearly is a complex scalar variable, so swap stuff around any way you like.

noah
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Consider an operator $f$ and a state \begin{equation} \psi(x) = \sum _n c_n \psi_n(x) \end{equation} with $\psi_n(x)$ being eigenfunctions of $f$. Then you get \begin{equation} f \psi(x) = \sum _n c_n f_n \psi_n \end{equation} According to your (statistical) formula \begin{equation} \langle f \rangle = \sum _n f_n P_n, \end{equation} where $P_n$ is the probability of finding the system in the state $\psi_n(x)$, which is given by \begin{equation} P_n = |c_n|^2. \end{equation} As a result you get \begin{equation} \langle f \rangle =\sum _n f_n |c_n|^2 = \int dx \, \psi^*(x)f\psi(x) \end{equation}

nwolijin
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In quantum mechanics, the expectation value of an observable $\hat{O}$ in a state $|\Psi\rangle$ is defined by $$ \langle \Psi|\hat{O}|\Psi\rangle \quad . $$

In your case the observable is the position operator with $\hat{x}\, |x\rangle= x\, |x\rangle$ and $\langle x|x'\rangle = \delta(x-x')$. We can write its expectation value, by making use of the relation $1 = \int \mathrm{d}x\, |x\rangle\langle x|$, as (here in 1D)

$$ \langle \Psi|\hat{x}|\Psi\rangle = \int\mathrm{d}x \int\mathrm{d}x'\, \langle \Psi|x\rangle\langle x| \hat{x}|x'\rangle\langle x'|\Psi\rangle \quad, $$

which reduces to $$\langle \Psi|\hat{x}|\Psi\rangle = \int \mathrm{d}x\, \Psi^*(x)\, x \, \Psi(x) =\int \mathrm{d}x\, x\, |\Psi(x)|^ 2 \quad ,$$ where we have defined $\langle x|\Psi\rangle \equiv \Psi(x)$.

Edit: As OP is asking also for the case of the momentum operator: We can make similar arguments here and find $$\langle \Psi|\hat{p}|\Psi\rangle = \int \mathrm{d}p\,p\, |\Psi(p)|^2\ \quad. $$

Edit 2: The above expectation value can also be carried out in the $x$-representation by using the fact that $\langle x|\hat{p}|\Psi\rangle = -i\hbar\, \partial_x \Psi(x)$: $$ \langle \Psi|\hat{p}|\Psi\rangle= -i\hbar \int \mathrm{d}x\, \Psi^*(x)\, \partial_x \Psi(x) \quad . $$

Tobias Fünke
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