# Understanding the calculation of expectation value

The expectation value (in sense of discrete probability) can be thought of as $$\left=\frac{1}{N}\sum\limits^{N}{Â }\psi$$ where $$N$$ is the number of experiments. As the number of experiments go to infinity the expression converts into an integral. $$\begin{array}{l} \left =\int_{-\infty }^{\infty }{{\psi }^{*}}Â\psi dx\\ \end{array}$$ The lower and upper bound are for $$\psi$$. Is the existence of $${\psi }^{*}$$ simply due to the integral of a complex function or is there more to it?

• Your first equation is simply wrong. You should have a $\psi^*$ in that equation to begin with. – Jahan Claes Feb 3 at 23:12
• @JahanClaes that's clearly a typo, since it's correct in the second equation – N. Steinle Feb 3 at 23:13
• @N.Steinle My interpretation of the question is that they are asking why the second equation has a $\psi^*$ and the first equation does not. It's a bit hard to parse, though. Hopefully if i'm wrong, OP edits for clarification. – Jahan Claes Feb 3 at 23:15
• The uppder and lower bounds are for the values of $x$, that is the spatial region over which you are making the statistical measurement of A. It is NOT over the values of $\psi$. – Bill N Feb 3 at 23:52
• @BillN ah yes. It's more specific and correct to say the upper and lower bounds are the range of $x$ over which $Â$ is applied on the function $\psi$. Just for interest, wouldn't this range also cover all the possible values of $\begin{array}{l} \Psi (x)\\ \end{array}$ since for an orthonomalized function $psi$ you have the following $\int_{-\infty }^{\infty }{{\Psi }^{*}}\Psi dx =1$ ? Hence, in this specific example, the upper and lower bound of $x$ is pretty much the same for $\begin{array}{l} \Psi (x)\\ \end{array}$ – Jung Feb 4 at 8:07

## 2 Answers

This is not correct in more than one way.

First, increasing the number of trials does not transform the sum into an integral. If you throw a die $$10^{20}$$ time the average value of the throw is still $$\frac{1}{10^{20}}\sum_n n N(n)$$ where $$N(n)$$ is the number of times you observe outcome $$n$$ on the die, i.e it still a sum over $$6$$ possible outcomes. More generally, $$\langle a\rangle =\frac{1}{N}\sum_a a A(a). \tag{1}$$ where $$A(a)$$ is the number of times you get the outcome $$a$$, and the sum is over all possible outcomes. The sum remains discrete even as you increase the number of experiments provided that the possible outcomes remain discrete, as in throwing a single die.

The key difference between something like Eq.(1) and $$\int dx P(x) x \tag{2}$$ is the nature of the outcomes, either discrete as in (1) or continuous like $$x$$ in (2).

In (2), the function $$P(x)$$ is a probably density, i.e.a non-negative function of $$x$$. Note that it is a density in the sense that, if $$\int dx P(x)=1$$, where the integration is over the entire range of $$x$$, the $$P(x_0)dx$$ is the probability of getting the $$x$$ values to lie between $$x_0\pm dx/2$$, i.e in the range $$x_0-dx/2$$ to $$x_0+dx/2$$. We use (2) to compute - say - the average position because the possible positions are a continuous set of outcomes.

The role of this non-negative probability density $$P(x)$$ is in quantum mechanics played by $$\psi^*(x)\psi(x)$$, which is also non-negative. For bound states we further require that $$\psi(x)^*\psi(x) <\infty$$ so we can rescale $$\psi(x)$$ so $$\int dx \psi^*(x)\psi(x)=1$$.

The expression you have involving $$\hat A\psi$$ is NOT an average value. In fact it’s just the state $$\psi$$ which results from the application of $$\hat A$$.

Is the existence of ψ∗ simply due to the integral of a complex function or is there more to it?

I shall use Dirac notation:

Given that the state of some quantum system is described by the ket, $$\lvert \psi \rangle$$, the wave function represented in position space is defined as $$\psi(x) = \langle x \lvert \psi \rangle$$, but I shall continue with just the abstract ket. We are guaranteed the existence of the complex transpose of $$\lvert \psi \rangle$$ by the fact that the Hilbert space that $$\lvert \psi \rangle$$ lives in is complex. More formally, the complex transpose of our state, $$\lvert \psi \rangle$$* = $$\langle \psi \rvert$$ is also called the dual of $$\lvert \psi \rangle$$ and is guaranteed to exist. This might also be a helpful conversation.

• thanks for telling me to look at the dual space of Hilbert space. But then the dual set of an infinite dimensional space (which is basically the Hilbert space in which $| \psi \rangle$ lives) does not span the dual space which is where $\langle \psi |$ lives in. Wouldnt that be a problem? – Jung Feb 4 at 18:26
• "The dual of an infinite-dimensional space has greater dimensionality (this being a greater infinite cardinality) than the original space has, and thus these cannot have a basis with the same indexing set. However, a dual set of vectors exists, which defines a subspace of the dual isomorphic to the original space. Further, for topological vector spaces, a continuous dual space can be defined, in which case a dual basis may exist. " en.wikipedia.org/wiki/Dual_basis#Existence_and_uniqueness – N. Steinle Feb 7 at 19:10
• I hope this discussion will help clear it up for you physics.stackexchange.com/questions/153178/… – N. Steinle Feb 7 at 19:25