# Expectation Values in Quantum Mechanics

Why is the expectation value what it is? Why don't you apply the operator, then multiply that by it's conjugate?

• So that I can understand the question better: could you clarify why you would expect applying the operator and then multiplying by its conjugate to give the expectation value? – David Z Sep 16 '13 at 0:17
• I just have no idea about the momentum operator. Trying to understand why the momentum operator should be given as an operator on a function, I'm now struggling to understand why you would do what you do for expectation values. I'm just generally confused. Showing that momentum is equivalent to a derivative is easy, but I don't understand why it's an operator, rather than just saying the product of the wave function and a momentum function yields the derivative. From here, my misunderstanding would have led me to think the operator should act on the normal, and conjugate wave function. – user24082 Sep 16 '13 at 0:25
• So, are you asking why we don't multiply the wavefunction $\psi(x)$ by some other function $f_p(x)$ to get the momentum? (Sorry, I'm still a little unclear on what you're confused about) Of course $\psi(x)f_p(x)$ would be a function, not a number, which is what we need momentum to be. – David Z Sep 16 '13 at 1:04
• I think he's asking why you don't do $\langle p \rangle = \int \psi^\dagger f_p(x) \psi$. After all, $\langle V \rangle = \int \psi^\dagger V(x) \psi$. Of course if you do it this way you don't get non-commutivity. – Brian Moths Sep 16 '13 at 1:22
• Can you write down an expression for "apply the operator, then multiply that by it's conjugate?" for example, do you mean $\langle p \rangle = |f_p(\psi(x,t))|^2$ for some $f_p$? – xuanji Sep 16 '13 at 2:07

Maybe this will help: at its core, expectation value is calculated the way it is because of basic probability. It has nothing to do with quantum mechanics. For example, if you have a random spinner or something (designated $X$) with three possible outcomes $A$, $B$, and $C$, and outcome $A$ has probability $P(A)$, outcome $B$ has probability $P(B)$, and outcome $C$ has probability $P(C)$, the expectation value is

$$\langle X\rangle = \frac{P(A) A + P(B) B + P(C) C}{P(A) + P(B) + P(C)}$$

The denominator is, of course, equal to one; I just put it to show how this is a weighted average. If you randomly produce a large number of results $X$ (e.g. spin the spinner a million times), the quantity $\langle X\rangle$ is the average of all those results.

This applies to quantum mechanics because we've observed (well, we assume, because it's consistent with observations) that quantum systems behave as random generators. For example, if you measure the momentum $p$ of a million identically prepared particles, the results you'll get will appear random. Of course, they do have some probability distribution $P(p)$ which depends on how you prepared the particles. That means that you can calculate the theoretical (or expected) average of those million momentum measurements like this:

$$\langle p\rangle = \frac{P(p_1)p_1 + P(p_2)p_2 + \cdots}{P(p_1) + P(p_2) + \cdots} = \sum_i P(p_i)p_i$$

or if the allowed momentum values form a continuous spectrum, just convert that into an integral:

$$\langle p\rangle = \int P(p)p\mathrm{d}p$$

But because the probability distribution over different momenta depends on how you prepared the particles (this is, again, something we can observe), you really should write the probability as a function of both the momentum $p$ and some nugget of information that describes the method of preparation, which I'll designate $\psi$:

$$\langle p\rangle_\psi = \frac{P(\psi, p_1)p_1 + P(\psi, p_2)p_2 + \cdots}{P(\psi, p_1) + P(\psi, p_2) + \cdots} = \sum_i P(\psi, p_i)p_i$$

or for continuous momenta,

$$\langle p\rangle_\psi = \int P(\psi, p)p\mathrm{d}p$$

All this is just to show that expectation values themselves are very simple, once you have the possible outcomes and their probabilities.

The part where quantum mechanics comes in, and the part that I think is confusing you, is how you get a probability out of a possible momentum $p$ and the information $\psi$ that describes how a particle was prepared. In other words, what is the nature of this function $P(\psi, p)$?

Well, one of the assumptions underlying quantum mechanics is that $\psi$ is an element of a Hilbert space. (We make this assumption because it results in a successful theory, not because it makes any sort of intuitive sense.) One of the defining characteristics of a Hilbert space is that it has an inner product $\langle f\lvert g\rangle$, which is something that takes two members of the space $\lvert f\rangle$ and $\lvert g\rangle$ and produces a number. Conveniently, this is very close to what we need $P(\psi, p)$ to do.

The logical way to try constructing $P(\psi, p)$ is to look for a fixed set of members of the Hilbert space, $\lvert p\rangle$, one for each momentum $p$, such that $P(\psi, p) = \langle\psi\lvert p\rangle$. This basically works, except that an inner product can produce a complex number, but that's easily fixed: just multiply it by its complex conjugate,

$$P(\psi, p) = \lvert\langle\psi\lvert p\rangle\rvert^2 = \langle\psi\lvert p\rangle\langle\psi\lvert p\rangle^* = \langle\psi\lvert p\rangle\langle p\lvert\psi\rangle$$

(Actually $\lvert p\rangle$ is not in the Hilbert space but that's a minor technicality.) This definition means that the expectation value can be expressed as

$$\langle p\rangle_\psi = \int \langle\psi\lvert p\rangle p\langle p\lvert\psi\rangle\mathrm{d}p = \langle\psi\lvert\biggl[\int \lvert p\rangle p\langle p\rvert\mathrm{d}p\biggr]\lvert\psi\rangle$$

Notice in the last piece of this I've broken the expression apart into a piece that describes how the particles are prepared, $\lvert\psi\rangle$ (and $\langle\psi\rvert$), and a piece that is independent of the specific particles being considered:

$$\int \lvert p\rangle p\langle p\rvert\mathrm{d}p$$

That latter piece specifies a recipe for taking two elements of the Hilbert space (in this example it's the same element twice, but it doesn't have to be) and turning them into a number. In this sense it's like the inner product, but it's more general than the inner product. Something which takes two elements of a space and produces a number is what physicists call an operator. (Mathematicians call it a bilinear form.) This is roughly why operators play such a central role in quantum mechanics: they can characterize a wide variety of ways to turn quantum states into numbers, which is essentially what we do when we measure a quantum system.

• How can you take the inner product beteeen two different Hilbert Spaces? – user24082 Sep 16 '13 at 15:59
• You can't, but why do you ask? I only use one Hilbert space in my answer. – David Z Sep 16 '13 at 16:46
• Momentum and wave functions belong the same Hilbert Space? – user24082 Sep 16 '13 at 17:41
• $\lvert p\rangle$ (or $\langle p\rangle$) isn't a momentum, it's a quantum state (you can think of it as just a wavefunction). The $p$ in $\langle p\rangle$ is just a label. Because there is one of these special states (wavefunctions) for each momentum, I use the momentum itself to label the state. But if it would make things clearer, I could change it to $\lvert\psi_p\rangle$ or something like that. – David Z Sep 16 '13 at 19:50
• And why did an additional $p$ show up in the expectation value formula's middle? – user24082 Sep 16 '13 at 21:44

I suppose, following a comment by @zodiac, that you would want, for the expectation value of an operator $A$ :

$$\langle A(t) \rangle = \int dx ~|(A (\psi(x,t))|^2 \tag{1}$$

This is, of course, totally wrong.

First, you may observe that the LHS term is linear in $A$, and that the RHS term is quadratic in $A$, so there is a clear incoherence, because in this case, we would have :

$\lambda \langle A(t) \rangle =\langle \lambda A(t) \rangle = \int dx ~|(\lambda A (\psi(x,t))|^2 = |\lambda|^2 \int dx ~|( A (\psi(x,t))|^2 = |\lambda|^2 \langle A(t) \rangle\tag{3}$

which is obviously false, for general $\lambda$.

The above remark shows us that the integral in the LHS term must be linear in $A$, the correct expression is :

$$\langle A(t) \rangle = \int dx~ \bar \psi(x,t)~ A ~\psi(x,t) \tag{4}$$