Determinate States

Assume that we have an ensemble of identical quantum systems all characterized by the state $|\psi \rangle$.

If we measure the value of the observable $Q$ in each of these systems will always find the same result? No, for an arbitrary quantum state $\psi \rangle$ each time we measure $Q$ we will find a different result.

Example

We have a spin $\frac{1}{2}$ system characterized by $|\psi \rangle = a |\uparrow \rangle + b |\downarrow \rangle$ and we let's measure its spin in the z-direction. Sometimes we will measure $+\frac{\hbar}{2}$ and for some other system we will measure $-\frac{\hbar}{2}$.

However, there are some special states where a measurement of $Q$ always returns the same result. So, we define determinate states for the observable $Q$ as those in which if we measure many times we find alwasy the same result.

Example

If $\hat{Q}=\hat{H}$, and if we have an ensemble of harmonic oscillators, all of them in the ground state, each time we measure the energy we will always find the same result, $E_0=\frac{\hbar \omega}{2}$. So, in this case the ground state is a determinate state.

Example

Now if the $\hat{Q}=\hat{S_z}$ and our quantum state is $|\psi \rangle = |\uparrow \rangle$ then this state corresponds to a determinate state of the $S_z$ since every measurement returns $+{\hbar}{2}$

In terms of operators, a determinate state obeys and eigenvalue equation: $\hat{Q}|\psi \rangle = q |\psi \rangle$. Let's illustrate this concept with an example.

Example

Let's consider a quantum system composed by only two energy levels:

$|\psi_1\rangle$ $\rightarrow$ $E_1$ and $|\psi_2\rangle$ $\rightarrow$ $E_2$

By definition, we know that $|\psi_1\rangle$ and $|\psi_2\rangle$ are determinate states of the total energy E, that is they are eigenstates of the Hamiltonian $\hat{H}$:

$$\hat{H}|\psi_1\rangle = E_1|\psi_1\rangle$$ $$\hat{H}|\psi_2\rangle = E_2|\psi_2\rangle$$

One can also construct valid quantum states which are not determinate, for instance: $\psi \rangle =\frac{1}{\sqrt{2}}|\psi_1\rangle-\frac{1}{\sqrt{2}}|\psi_2\rangle$. If we apply the $\hat{H}$ to this quantum state we find $$\hat{H}|\psi \rangle=\hat{H}(\frac{1}{\sqrt{2}}|\psi_1\rangle-\frac{1}{\sqrt{2}}\psi_2\rangle) = \frac{E_1}{\sqrt{2}}|\psi_1\rangle-\frac{E_2}{\sqrt{2}}|\psi_2\rangle \neq E |\psi \rangle$$ . So, $|\psi \rangle$ is not a determinate state.

Now we know that determinate states are eigenstates of the Hermitian operator.

In the following we will see a number of important properties of the eigenvalues and eigenfunctions of the Hermitian operators.

Let's consider an eigenvalue equation for an Hermitian operator:

$$\hat{Q}|\psi \rangle = q |\psi \rangle$$

As we have seen so far, the spectrum could be discrete (i.e. harmonic oscillator), continuous (i.e. the free particle), or a combination of a discrete and continuos (i.e. the delta fucntion and the finite potential well).

Let's discuss some relevant properties for the discrete and continuous cases:

For the discrete case:

1. The eigenvalues are real ($q \in \mathbb{R}$).

   Proof: Let's use the definition of an Hermitian operator, $\hat{Q}=\hat{Q}^{\dagger}$. $$\langle \psi |\hat{Q}|\psi \rangle = \langle \psi |\hat{Q} \psi \rangle = q \langle \psi | \psi \rangle$$ $$\langle \psi |\hat{Q}|\psi \rangle = \langle \psi \hat{Q}| \psi \rangle = q^* \langle \psi | \psi \rangle$$ $$q \langle \psi | \psi \rangle = q^* \langle \psi | \psi \rangle$$ therefore if a number is equal to its complex conjugate the number $q \in \mathbb{R}$.

2. The eigenfunctions associated to different eigenvalues are orthogonal among them:

   $$\int_{-\infty}^{+\infty}\psi^*_n \psi_l=\delta_{nl}$$

   Proof: Let's consider two eigenfunctions of $\hat{Q}$ with different eigenvalues: $$\hat{Q}|\psi_1 \rangle = q_1 |\psi_1 \rangle$$ and $$\hat{Q}|\psi_2 \rangle = q_2 |\psi_2 \rangle$$ Since $\hat{Q}$ is Hermitian operator we have that, $\langle \psi_1 |\hat{Q}\psi_2 \rangle=\langle \psi_1 \hat{Q}|\psi_2 \rangle$, and this implies, $$q_2\langle \psi_1 |\psi_2 \rangle=q_1^*\langle \psi_1|\psi_2 \rangle$$ But since we know that $q_1$ and $q_2$ are real, then the only way this equality is satistfied is if $\langle \psi_1|\psi_2 \rangle=0$ Therefore, $|\psi_1 \rangle$ and $|\psi_2 \rangle$ are orthogonal.

3. The eigenstates of the operator $\hat{Q}$ associated to the observable $Q$ are complete. That is, any function in Hilbert space can be expressed as a linear combination of these eigenvalues: $\hat{Q}|\psi_n \rangle = q_n |\psi_n \rangle$, $|\Psi \rangle = \sum_n c_n |\psi_n \rangle$

For the continuous case:

Important

In the continuous case the proofs of the properties of the discrete case do not apply, since the eigenstates are not normalizable and the inner product might not exist. So, we need a different way to show the same properties: 1.The eigenvalues are real. 2.The eigenstates are orthogonal. 3.The eigenvalues form a complete basis.

We will show properties 1, 2, and 3 via examples.

The eigenvales are real and orthogonal:

Say that we want to find the eigenvalues and eigenstates of the momentum operator $\hat{p}$. In the following we will work in the position representation, where $f_p(x)$ is the eigenfunction of the linear momentum operator.

$$\hat{p}f_p(x)=pf_p(x)$$

in this case the eigenvalue equation that we need to solve is:

$$\frac{\hbar}{i}\frac{\partial}{\partial x} f_p(x)=p f_p(x)$$

the general solution is:

$$f_p(x)=Ae^{i\frac{px}{\hbar}}$$

which is not normalizable.

Despite $f_p(x)$ is not normalizable, let's try to calculate the inner product of two different eigenfunctions: $f_p_1(x)$ and $f_p_2(x)$.

$$\langle f_p_1(x)|f_p_2(x) \rangle = \int_{-\infty}^{+\infty} dx |A|^2 e^{i\frac{(p_1-p_2)x}{\hbar}}=|A|^2\int_{-\infty}^{+\infty} dx e^{i\frac{(p_2-p_1)x}{\hbar}}=$$ we assume that the eigenvalues $p_1$ and $p_2$ are real (we are not going to demonstrate it)

$$=2\pi \hbar |A|^2 \delta (p_2-p_1)$$

We chose $A=\frac{1}{\sqrt{2 \pi \hbar}}$ and we find that the eigenfunctions of the momentum operator $\hat{p}$ are,

$$f_p(x)=\frac{1}{\sqrt{2 \pi \hbar}}e^{i\frac{px}{\hbar}}$$ and satifsy the following relation,

$$\langle f_p_1(x)|f_p_2(x) \rangle =\delta (p_2-p_1)$$ or

$$\langle f_p_1(x)|f_p_2(x) \rangle =\delta (p_1-p_2)$$ because the Delta function is symmetric.

If $p_1 \neq p_2$ then $\delta (p_2-p_1)=0$, which is teh analogue in the continuum case than the orthogonality relation in the discrete case.

Note that, this relation $\langle f_p_1(x)|f_p_2(x) \rangle =\delta (p_2-p_1)$ is called Dirac orthogonality.

The eigenfunctions form a complete basis:

$\psi(x)=\int_{-\infty}^{+\infty} dp c(p) f_p(x)=\frac{1}{\sqrt{2}}\int_{-\infty}^{+\infty} dp c(p) e^{i\frac{px}{\hbar}}$