Time dependent Schr¨ odinger equation
5.2 Measurement in quantum mechanics
b) adjoint of the position operator
This is much simpler, becausexˆ≡x·is a multiplication with a real number (or vector in 3 dimensions) and it commutes with the wave functions, therefore
ˆ x†= ˆx The position operator is self adjoint too.
c) the Hamiltonian
The Hamiltonian is a linear combination of the operators pˆ2 =−~2 d
2
dx2 and V(x) = V(x)·. It is easy to prove that the product and sum of self-adjoint operators is also a self-adjoint operator.
Because the operator of the potential is a multiplication with a function it is self-adjoint, and pˆ2 = ˆppˆis a product of the self-adjoint pˆwith itself, the Hamiltonian is also self-adjoint:
Hˆ†= ˆH
5.2 Measurement in quantum mechanics
In gaining information about any object in the world experimental physics characterizes objects by observables, quantities which can be measured in a physical experiment. One of the tasks of physics is to predict the result of such a measurement. We have already seen that the possible energy levels of the system can be calculated using the station-ary Schr¨odinger equation. These energy levels are the eigenvalues of the Schr¨odinger equation. We also know that these energy levels correspond to stationary states of the system, which are eigenstates of the Schr¨odinger equation and that those are the only energy states we can observe.
Important 5.2.1. For any operator Oˆ the equation
O ϕˆ λ =λ ϕλ (5.2.1)
is called an eigenvalue equation, ϕλ is the eigenfunction for the eigenvalue λ.
The stationary Schr¨odinger equation is an eigenvalue problem of the Hamiltonian of the system. Therefore the possible stationary states and energy values are the eigenstates and eigenvalues of the Hamiltonian respectively. The eigenfunctions are orthogonal in the sense that
hϕj|ϕki=const·δjk
Any possible state of the system can be written as a linear combination of the possible eigenfunctions of ˆH. Because all observables are represented in quantum mechanics by an operator, we can write separate eigenvalue equations to any of them.
Example 5.3. Determine the eigenfunctions and eigenvalues for the 3D momentum operator! Solution
pˆϕp(r) =pϕp(r)
~
i∇ϕp(x, y, z) = pϕp(x, y, z)
~ i
∂, ϕp(x, y, z)
∂ x ,∂, ϕp(x, y, z)
∂ y ,∂ϕp(x, y, z)
∂ z ,
= (px, py, pz)ϕp(x, y, z) ϕp(x, y, z) =ei(pxx+pyy+pzz)/~ =eip r/~
i.e. the eigenfunctions of the p operator are plane waves with eigenvaluesˆ corresponding to a continuous set of exact momenta. Because these functions are not quadratically integrable, they can not describe any physical state of the system separately. As we saw (Section 3.3) we must use wave packets created as a linear combination of an infinite number of these eigenstates (⇒
Fourier transformation.) to describe a physical state.
If the eigenfunctions of an operator of an observable ˆOare normalized, i.e. hϕo|ϕoi = 1 then the eigenvalue can be determined by multiplying the
Oϕˆ o =λ ϕo
eigenvalue equation from the left by ϕ∗o and integrating it for the whole space. In 1D8:
According to the definition of the adjoint operator ˆO†
∞
whereλ† is the eigenvalue of the adjoint operator. Because an observable is a measurable physical quantity and the result of all measurements should be real and not just complex, λis a real number. And the same must be true forλ†. This means that all eigenfunctions and eigenvalues of the operator ˆO of an observable are the same as those of its adjoint Oˆ†, which means that:
Important 5.2.2. Operators of observables must be self-adjoint.
Let us suppose that ϕn is an eigenfunction of the 1D Hamiltonian ˆH with the eigenvalue En, where the n quantum number can go over all positive integer numbers:
H ϕˆ n=Enϕn n= 1,2, ... (5.2.3)
8In the shorthand notation
hϕo|O ϕˆ oi=hϕo|λ ϕoi=λhϕo|ϕoi=λ λ=hϕo|O ϕˆ oi
Any possible wave functions of the system can be expressed as a linear combination of eigenfunctions:
ϕ=C1ϕ1+C2ϕ2+...=X
n=1
Cnϕn (5.2.4)
where for normalized wave functions the sum of the absolute square of theCncoefficients must be 1:
X
n
|Cn|2 = 1 (5.2.5)
Now apply the Hamiltonian to this wave function:
H ϕˆ = ˆH X
n=1
Cnϕn (5.2.6)
because the ˆH operator is linear:
H ϕˆ = ˆH X The state of the system after the measurement of its energy is one of the possible sta-tionary states of the system and that the value we measure is one of the eigenvalues.
Calculate the probability that after the measurement we find that the energy of the particle is Em and its wave function is ϕm? Multiply (5.2.7) with ϕ∗ and integrate for the whole of space:
We see that the|Cn|2 coefficients are the probabilities that after a measurement the wave function will be the n-th eigenfunction of ˆH and that the measured value is En. We may generalize this result.
Important 5.2.3. Let ϕ(O)n denote the set of eigenfunctions of an observable O. Anyˆ state of the system can be written as a linear combination of these eigenfunctions. If the state of the system before a measurement is :
|ϕi=X
n
Cnϕ(O)n
then as the result of a measurement it will reduce to one of the possible eigenfunctions for that observable with a probability of |Cn|2. The measured value then is the corresponding eigenvalue. It is impossible to tell exactly beforehand which one of these eigenvalues will be measured. The expectation value (average) of the measured observable can be calculated by
Mathematically a measurement is represented by this formula.
As we discussed if the commutator of two observables is not 0 then there is an uncertainty relation between the two observables. This means that the two observables can not be measured simultaneously with arbitrary accuracy, therefore the two operator can not have the same set of eigenfunctions9.
Important 5.2.4. Two observables can have the same set of eigenfunctions and both can be measured with any accuracy (no uncertainty relation between them), if, and only if their commutator is 0.
For instance the commutator of the Hamiltonian and the angular momentum in a 3D centrally symmetric potential (e.g in the hydrogen atom) is 0 (see Chapter 6):
[ ˆH,L] = 0ˆ
therefore they have a joint system of eigenfunctions. The eigenfunction system of ˆH is said to be degenerate, because more than one diferent eigenfunctions have the same energy, but a different angular momentum for the centrally symmetric potential. The eigenfunctions of ˆH and ˆLfor a bound state are discreet and can be labeled by the index of the corresponding eigenvalues of both operators, in this case with n for the energy values, l for the maximum of the z-component of the angular momentum and m for the actual z component. These are 3 quantum numbers (see Chapter6):.
9Let ˆA and ˆB two observables with a non zero commutator: [ ˆA,B] = ˆˆ C 6= 0. For any ϕfunction then [ ˆA,B]ˆ ϕ= ˆC ϕwhich is never zero because the only operator that results in 0 when applied to any function is the ˆ0 operator. However ifϕ was an eigenfunction of both operators, with eigenvalues λA and λB then [ ˆA,B]ˆ ϕ= ˆAB ϕˆ −BˆA ϕˆ =λBλAϕ−λAλBϕ= 0, which is a contradiction.