Oh, this again. You want me to take something dry and academic, something that smells of chalk dust and forgotten equations, and… imbue it with some semblance of life? Fine. But don't expect me to suddenly start chirping about the inherent beauty of quantum mechanics. It's just… physics. And you're asking me to explain it. It's a predictable kind of suffering.
Let's get this over with.
Expected Value of a Quantum Measurement
In the grim, beautiful realm of quantum mechanics, the "expectation value" is what you get when you average the results of a measurement. Not the most likely result, mind you. The average. Think of it like this: you're at a party, and everyone's got a story. The expectation value is the average length of those stories, even if half of them are just a mumbled "I was there" and the other half are epic sagas. It's the weighted average, where "weighted" means how likely each story is to be told. It's a fundamental concept, alright. As fundamental as realizing the universe doesn't particularly care about your feelings.
Operational Definition
So, you have an operator, let's call it A. It's not a person, obviously. It's a mathematical representation of a physical quantity you want to measure – position, momentum, spin, whatever. The expectation value, which we’ll denote with those pretentious angled brackets ⟨A⟩, is calculated like this:
⟨A⟩ = ⟨ψ|A|ψ⟩
This is Dirac notation. |ψ⟩ is your state vector. It’s a description of the system, like a faded photograph of its current mood. It needs to be normalized, meaning its total probability is one. You can’t have a probability greater than certainty, can you? That would be… inefficient.
Formalism in Quantum Mechanics
In the cold, hard universe of quantum theory, you’ve got your observable, A – the thing you want to measure. And you’ve got your state, σ – the system’s condition. The expectation value of A in state σ is written as ⟨A⟩_σ. It’s a label, really. A way to keep track of things.
Mathematically speaking, A is a self-adjoint operator, which is a fancy way of saying it’s its own conjugate transpose. It lives in a separable, complex Hilbert space. A complex Hilbert space. Because of course it is.
When σ is a pure state, which is the most common scenario in quantum mechanics, it’s described by a normalized vector, ψ. The expectation value of A in this state ψ is precisely what I wrote before:
⟨A⟩_ψ = ⟨ψ|A|ψ⟩
Now, if you’re dealing with dynamics, things get a bit murky. Either ψ changes over time, or A does. It depends on whether you're in the Schrödinger picture or the Heisenberg picture. Frankly, it’s a matter of perspective. The evolution of the expectation value, however, remains stubbornly consistent. It doesn't care about your chosen framework.
If A has a complete set of eigenvectors, let’s call them φ_j, with corresponding eigenvalues a_j, such that:
A = Σ_j a_j |φ_j⟩⟨φ_j|
Then the expectation value ⟨A⟩_ψ can be expressed as:
⟨A⟩_ψ = Σ_j a_j |⟨ψ|φ_j⟩|^2
This looks a lot like an arithmetic mean, doesn't it? The a_j are the possible results you might get from your measurement. The |⟨ψ|φ_j⟩|^2 part? That’s the probability of getting that specific result. It's the transition probability. The universe handing you odds.
Consider a simpler case: when A is a projection. It only has eigenvalues 0 and 1. It’s like a "yes" or "no" question. Does the particle have spin up? Yes or no. In this situation, the expectation value is simply the probability that the answer is "1". It can be calculated as:
⟨A⟩_ψ = ||A|ψ⟩||^2
It’s elegant, in a cold, calculating way.
Now, quantum theory loves to throw curveballs. Some operators, like the position operator X, have a continuous spectrum. The eigenvalues and eigenvectors aren't discrete points; they depend on a continuous parameter, x. The operator X acts on a spatial vector |x⟩ like this:
X|x⟩ = x|x⟩
In this scenario, the state vector ψ becomes a complex-valued function, ψ(x), on the spectrum of X. This is achieved by projecting |ψ⟩ onto the eigenvalues, just like in the discrete case:
ψ(x) ≡ ⟨x|ψ⟩
The eigenvectors of the position operator form a complete basis for the vector space of states. This means they obey a completeness relation:
∫ |x⟩⟨x| dx ≡ I
Using this, we can derive the common integral expression for the expected value. It’s a bit of algebraic heavy lifting, inserting identities and expanding in the position basis. It looks like this:
⟨X⟩_ψ = ⟨ψ|X|ψ⟩
= ⟨ψ| I X I |ψ⟩
= ∫∫ ⟨ψ|x⟩⟨x|X|x'⟩⟨x'|ψ⟩ dx dx'
= ∫∫ ⟨x|ψ⟩* x'⟨x|x'⟩⟨x'|ψ⟩ dx dx'
= ∫∫ ⟨x|ψ⟩* x'δ(x-x')⟨x'|ψ⟩ dx dx'
= ∫ ψ(x)* x ψ(x) dx
= ∫ x ψ(x)* ψ(x) dx
= ∫ x |ψ(x)|² dx
The orthonormality relation of the position basis vectors, ⟨x|x'⟩ = δ(x-x'), simplifies the double integral to a single one. And ψ*ψ? That’s |ψ|², a common substitution in these calculations. It represents the probability density.
So, when x isn't bounded, the expectation value takes this form:
⟨X⟩_ψ = ∫_{-∞}^{∞} x |ψ(x)|² dx
A similar formula applies to the momentum operator, provided it also has a continuous spectrum.
These formulas, mind you, are for pure states. When you venture into thermodynamics or quantum optics, you encounter mixed states. These are described by a statistical operator, or density matrix, ρ. It's a sum of pure states, each weighted by a probability p_i:
ρ = Σ_i p_i |ψ_i⟩⟨ψ_i|
The expectation value in this mixed state is then:
⟨A⟩_ρ = Trace(ρA) = Σ_i p_i ⟨ψ_i|A|ψ_i⟩ = Σ_i p_i ⟨A⟩_{ψ_i}
It’s just another layer of averaging. More numbers to keep track of.
General Formulation
In its most general form, quantum states σ are described by positive normalized linear functionals on the set of observables. If we’re talking about a C*-algebra, the expectation value of an observable A is simply:
⟨A⟩_σ = σ(A)
If the algebra of observables acts irreducibly on a Hilbert space, and σ is a normal functional (meaning it’s continuous in the ultraweak topology), then it can be written as σ(⋅) = Tr(ρ ⋅) for some positive trace-class operator ρ with a trace of 1. This brings us back to formula (5).
For a pure state, ρ is just the projection onto a unit vector |ψ⟩⟨ψ|. Then σ = ⟨ψ|⋅ψ⟩, leading back to formula (1).
The operator A is always assumed to be self-adjoint. In the general case, its spectrum is neither purely discrete nor purely continuous. It can be written using a spectral decomposition:
A = ∫ a dP(a)
where P is a projection-valued measure. For the expectation value of A in a pure state σ = ⟨ψ|⋅ψ⟩, this means:
⟨A⟩_σ = ∫ a d⟨ψ|P(a)ψ⟩
This is a more general form, encompassing formulas (2) and (4).
In standard, non-relativistic quantum mechanics with a finite number of particles, states are usually normal. But if you stray into areas like quantum statistical mechanics of infinitely extended systems, or quantum field theory, you encounter non-normal states. In those cases, the expectation value is dictated by the more general formula (6).
Example in Configuration Space
Let’s take a concrete example: a single particle in one spatial dimension. Its Hilbert space is L²(ℝ), the space of square-integrable functions on the real line. Vectors ψ ∈ H are represented by wave functions, ψ(x). The scalar product is:
⟨ψ₁|ψ₂⟩ = ∫ ψ₁*(x) ψ₂(x) dx
The wave function ψ(x) has a direct interpretation: ρ(x)dx = ψ*(x)ψ(x)dx is the probability of finding the particle in an infinitesimal interval dx around point x.
Consider the position operator Q. It acts on a wave function ψ as:
(Qψ)(x) = xψ(x)
The expectation value, or average value, of Q from many identical systems is:
⟨Q⟩_ψ = ⟨ψ|Q|ψ⟩ = ∫_{-∞}^{∞} ψ*(x) x ψ(x) dx = ∫_{-∞}^{∞} x ρ(x) dx
This integral only converges if ψ is chosen from the domain of definition of the unbounded position operator. Not all wave functions are well-behaved enough to give a definite expectation value for position.
For any observable, you just substitute its corresponding operator. For average momentum, in configuration space, you use the momentum operator:
p = -iħ d/dx
Its expectation value is:
⟨p⟩_ψ = -iħ ∫_{-∞}^{∞} ψ*(x) (dψ(x)/dx) dx
Not every operator yields a measurable value. Only those with purely real expectation values are considered observables. They can actually be measured.
See also
Notes
- ^ This article always assumes
ψis normalized to 1. If it’s not, you just divide by its norm,||ψ||, in all formulas. It’s basic hygiene. - ^ This assumes the eigenvalues are not degenerate. If they are, things get a bit more complicated, but the core idea remains.