The Meaning of Basis States in Quantum Mechanics

[BigMax]

Hi everyone!

I've been studying quantum mechanics for a while but I have a big big problem. If a system is in an eigenstate of energy (I use the eigenstate as a basis) it remains in this state forever. But if I describe the system with a different set of basis states (not eigenstates) the system can be found in AN energy eigenstate with some probability and than it can move to an another state. How is it possible?

Max

 

[A. Neumaier]

If a system is in an eigenstate of energy (I use the eigenstate as a basis) it remains in this state forever.

Only if it is completely isolated. To observe (''find'') it, one must break the isolation.

 

[Nugatory]

I've been studying quantum mechanics for a while but I have a big big problem. If a system is in an eigenstate of energy (I use the eigenstate as a basis) it remains in this state forever. But if I describe the system with a different set of basis states (not eigenstates) the system can be found in AN energy eigenstate with some probability and than it can move to an another state. How is it possible?

Can you give an example of the situation you're considering? There are several different cases that might be described that way.

 

[andresB]

Hi everyone!

I've been studying quantum mechanics for a while but I have a big big problem. If a system is in an eigenstate of energy (I use the eigenstate as a basis) it remains in this state forever. But if I describe the system with a different set of basis states (not eigenstates) the system can be found in AN energy eigenstate with some probability and than it can move to an another state. How is it possible?

Max

It is not possible.

If you expand your energy eigenstate into another basis, the probability to measure that energy eigenvalue (and not some other energy eigenvalue) doesn't change.

 

[andresB]

By the way, you can't use an eigenstate as a basis, because an eigenstate is just one vector. The basis is formed by all the energy eigenstates.

In general, a linear combination of energy eigenstates (corresponding to different energy eigenvalues) is not stationary.

 

[BigMax]

Can you give an example of the situation you're considering? There are several different cases that might be described that way.

For example I've been studying systen with two basis states. I choose one basis and then I fondi the eigenstate which form another bases. There is AN amplitude for finding my bases states in the energy eigenstate. What does it mean if the system can be described with rhe tuo states I've choosen? If the system is a tuo basis states system it can only move between the tuo states I chose, right?

 

[vanhees71]

First of all you have to get clear about the concepts. The (pure) states of a quantum system are described by a "state vector" (more precisely by a state vector modulo an arbitrary non-zero factor, a socalled ray, but that are finer details). In the socalled Schrödinger picture of time evolution the state vector $|\psi(t) \rangle$ is a function of time.

Observables are described by self-adjoint operators, and there is a particular operator, the Hamilton operator (usually representing the total energy of the system) which determines the time evolution of the state by the equations
$$\mathrm{i} \partial_t |\psi(t) \rangle=\hat{H} |\psi(t) \rangle.$$
If $\hat{H}$ is not explicitly time dependent a formal solution is
$$|\psi(t) \rangle=\exp \left (-\frac{\mathrm{i}}{\hbar} \hat{H} t \right ) |\psi(0) \rangle.$$
The values an observable $A$ can take when precisely measured are the eigenvalues of the corresponding self-adjoint operator $\hat{A}$. Assuming that for any eigenvalue $a$ there is only one eigenvector (modulo a non-zero factor of course), $|a \rangle$ then the probability to get the value $a$ when measuring the observable at time $t$, given the state $|\psi(t) \rangle$ is
$$P(t,a)=|\langle a|\psi(t)\rangle|^2.$$
The factors for both vectors have to be chosen such that they are normalized, i.e., $\langle \psi|\psi \rangle=\langle a|a\rangle=1$.

Usually the eigenvectors of a self-adjoint operator form a complete orthonormal set (or a basis), i.e., you can expand any vector wrt. this basis of eigenvectors.
$$|\psi(t) \rangle=\sum_a |a \rangle \langle a|\psi(t) \rangle.$$
Now to solve the equation of motion given above (which is the Schrödinger equation in Dirac's bra-ket notation) a particularly convenient basis is the energy eigenbasis,
$$\hat{H} |E \rangle=E |E \rangle.$$
So we write
$$|\psi(t) \rangle=\sum_E |E \rangle \langle E|\psi(t) \rangle.$$
Now we have
$$|\psi(t) \rangle = \exp \left (-\frac{\mathrm{i}}{\hbar} \hat{H} t \right) |\psi(0) \rangle.$$
Plugging our decomposition in terms of energy eigenvectors in we simply get
$$|\psi(t) \rangle = \sum_E \exp \left (-\frac{\mathrm{i}}{\hbar} E t \right) |E \rangle \langle E |\psi(0) \rangle.$$
Particularly, if at time $t=0$ the state vector is an eigenvector (i.e., the system is prepared in an energy eigenstate) $|\psi(0) \rangle=|E_0 \rangle$ the equation simplifies to
$$|\psi(t) \rangle = \exp \left (-\frac{\mathrm{i}}{\hbar} E_0 t \right) |E_0 \rangle.$$
All that happens to the initial state vector is that it gets multiplied by a phase factor, but this doesn't really change the state, because all statevectors that only differ by just some non-zero factor describe the same (pure) state. This shows that the energy eigen states are the stationary states of the system. This distinguishes the energy eigenstates from any other basis.

Of course, often you have to describe the physics in terms of other bases. E.g., if you learn QT in terms of "wave mechanics" you start working in the position basis. Since the position operator $\hat{x}$ has continuous eigenvalues $x \in \mathbb{R}$ instead of sums you have to write integrals. Thus in this case the state vector is written as
$$\psi(t) \rangle=\int_{\mathbb{R}} \mathrm{d} x |x \rangle \langle x|\psi(t) \rangle,$$
and
$$\psi(t,x)=\langle x|\psi(t) \rangle$$
is the wave function. Now the Schrödinger equation is expressed for the wave function as
$$\mathrm{i} \hbar \partial_t \psi(t,x)=\hat{H} \psi(t,x),$$
where now $\hat{H}$ acts on the wave function. E.g., for a particle moving in some potential the Hamiltonian is indeed given as the total energy operator,
$$\hat{H}=\frac{1}{2m} \hat{p}^2+V(x)=-\frac{\hbar^2}{2m} \partial_x^2 + V(x).$$
From this you can evaluate the energy eigenvalues and eigenvectors by solving the eigenvalue equation (also known as the "time-independent Schrödinger equation")
$$\hat{H} u_E(x)=E u_E(x).$$
In Dirac notation
$$u_E(x)=\langle x|E \rangle.$$
The formal solution of the time-dependent Schrödinger equation then follows immediately:
$$\psi(t,x)=\langle x|\psi(t) \rangle=\sum_E \exp \left (-\frac{\mathrm{i}}{\hbar} t E \right ) \langle x|E \rangle \langle E|\psi(0) \rangle =\sum_E \exp \left (-\frac{\mathrm{i}}{\hbar} t E \right ) u_E(x) \psi_{E0},$$
where
$$\psi_{E0}=\langle E|\psi(0) \rangle = \int_{\mathbb{R}} \mathrm{d} x \langle E|x \rangle \langle x|\psi(0) \rangle = \int_{\mathbb{R}} \mathrm{d} x u_E^*(x) \psi_0(x).$$

 

[hilbert2]

The only way I know how to create a stationary state ($\partial_t |\Psi(x,t)|^2 = 0$ for all $x,t$) that is not an eigenstate of $\hat{H}$ is to put two "boxes" of different length on the x-axis so that the $V(x)=0$ whenever $x_1 < x < x_2$ or $x_3 < x < x_4$ and $V(x)=\infty$ otherwise. Here $x_1 < x_2 < x_3 < x_4$ and $x_2 - x_1 \neq x_4 -x_3$. Now if you make a wave function that is a sum of the ground states of each individual box, the probability density at any point $x$ is stationary but the system is not an eigenstate of $\hat{H}$ unless the ground state energies of the two boxes are equal. This is caused simply by the infinite potential barrier preventing any tunneling between the boxes. I guess this is what A. Neumaier meant with a system being in complete isolation. However, infinite barriers don't really exist in the real world.

 

[Nugatory]

What does it mean if the system can be described with rhe tuo states I've choosen? If the system is a tuo basis states system it can only move between the tuo states I chose, right?

Changing the basis does not change the state, it just changes the way it is represented mathematically. It's like how we might describe a direction using the NS/EW basis or the forwards/sideways basis: when I'm facing east "two steps forward, one step left" and "one step north, two steps east" are different ways of describing the same direction.

The system state does not have to be one of the basis states; any vector sum of the bases is possible. It's only when we measure some observable that the system state collapses to some eigenfunction of that observable (and the measurement result is the corresponding eigenvalue). Thus it is convenient to choose as a basis the eigenvectors of whatever observable we're interested in.