I Feynman's description of the quantum behaviour of an ammonia molecule

[Roby002]

Hi guys! These days I've been reading Feynman's description of the quantum behaviour of an ammonia molecule. He assumes the N up or down as tuo basis states. He then says there's a little probability that the state UP becomes DOWN and viceversa. But for definition basis states are orthogonal...so how the UP state can become a DOWN one?

 

[hilbert2]

If two basis states are not exact eigenstates of the Hamiltonian, but are close to being that, then transitions between those basis states happen slowly and with small probability.

 

[Roby002]

But could the be used as basis states?

 

[Haborix]

You can always pick any basis states you like. The physics which happens with those states might suggest a preference for using one basis over another. But in the end, nothing changes by using a different choice of basis states, aside from our ability to perceive and interpret the results of the calculations.

 

[Roby002]

But why Feynaman says that one should use orthogonal states? And then he doesen't use them for the ammonia?

 

[PeterDonis]

You can always pick any basis states you like.

Not quite. You have to pick a set of basis states that are linearly independent and that span the space. Not all sets of states meet those requirements.

But why Feynaman says that one should use orthogonal states

Where does he say that?

And then he doesen't use them for the ammonia?

Because basis states don't have to be orthogonal. They just have to be linearly independent, which is a weaker condition.

 

[Haborix]

Not quite. You have to pick a set of basis states that are linearly independent and that span the space. Not all sets of states meet those requirements.

Thanks, that's probably a good clarification to make in the context of this thread. As a matter of nomenclature, I have always understood "basis states/vectors" to mean a set of vectors which is linearly independent and spanning.

 

[hilbert2]

Two basis states being orthogonal doesn't prevent transitions between them, unless both are eigenstates of the Hamiltonian operator of the system.

This problem is similar to the symmetric quantum 1D double well, which has two compartments and a potential barrier in between. There the two lowest eigenstates of the Hamiltonian have almost the same energy (if the potential barrier is high enough), but the ground state wave function is symmetric about the midpoint of the system and the 1st excited state is antisymmetric. Now you can make the linear combinations $2^{-1/2}(\left|\psi_1 \rangle\right. + \left|\psi_2 \rangle)\right.$ and $2^{-1/2}(\left|\psi_1 \rangle\right. - \left|\psi_2 \rangle)\right.$ from the two lowest energy eigenstates $\left|\psi_1 \rangle\right.$ and $\left|\psi_2 \rangle\right.$. These linear combinations describe a situation where the particle is only on one side of the double well, and there exists a small probability of transitions between them because the $\left|\psi_1 \rangle\right.$ and $\left|\psi_2 \rangle\right.$ don't have exactly the same energy and therefore their linear combination is not an exact eigenstate of $\hat{H}$.

 

[physicsworks]

But for definition basis states are orthogonal...so how the UP state can become a DOWN one?

With the passage of time, of course. The fact that UP and DOWN states ($|1 \rangle$ and $|2 \rangle$ in Feynman's notation) are orthogonal does not mean that if you start in, say, state $|1 \rangle$ you cannot possibly end up in state $|2 \rangle$. It is the unitary time evolution governed by the corresponding hamiltonian of the ammonia molecule that dictates how the initial state evolves in time. The energy eigenstates of the hamiltonian are what Feynman calls $| I \rangle = \frac{1}{\sqrt{2}} (|1 \rangle - |2 \rangle)$ and $| II \rangle = \frac{1}{\sqrt{2}} (|1 \rangle + |2 \rangle)$, they correspond to energies $E_{I, II} = E_0 \pm A$, respectively. These energy eigenstates are what we call stationary states; for them, time evolution is trivial and amounts to just picking up a phase factor (it is trivial because the hamiltonian is time independent). It is for these energy eigenstates that the statement "if one preapres the system in the state $| I \rangle$ at $t=0$, then after some time $t$ one finds the system in the state which is the initial one times the phase factor $\exp(-i E_{I} t/\hbar)$, and similarly for $| II \rangle$" is true. These energy eigenstates also constitute a basis, so any initial state, e.g. $|1 \rangle$, can be written as a linear combination of them: $|1 \rangle = \frac{1}{\sqrt{2}} (| I \rangle + | II \rangle$).

Thus to determine time evolution of the initial state one can:
1) expand the initial state in the basis of energy eigenstates;
2) time-evolve the expanded state using the fact that each energy eigenstate picks up a phase factor with the corresponding energy eigenvalue in the exponent;
3) go back to the original basis;
4) interpret coefficients in front of the original basis vectors as probability amplitudes of fidning the system at a later time in those states.

 

[Roby002]

With the passage of time, of course. The fact that UP and DOWN states ($|1 \rangle$ and $|2 \rangle$ in Feynman's notation) are orthogonal does not mean that if you start in, say, state $|1 \rangle$ you cannot possibly end up in state $|2 \rangle$. It is the unitary time evolution governed by the corresponding hamiltonian of the ammonia molecule that dictates how the initial state evolves in time. The energy eigenstates of the hamiltonian are what Feynman calls $| I \rangle = \frac{1}{\sqrt{2}} (|1 \rangle - |2 \rangle)$ and $| II \rangle = \frac{1}{\sqrt{2}} (|1 \rangle + |2 \rangle)$, they correspond to energies $E_{I, II} = E_0 \pm A$, respectively. These energy eigenstates are what we call stationary states; for them, time evolution is trivial and amounts to just picking up a phase factor (it is trivial because the hamiltonian is time independent). It is for these energy eigenstates that the statement "if one preapres the system in the state $| I \rangle$ at $t=0$, then after some time $t$ one finds the system in the state which is the initial one times the phase factor $\exp(-i E_{I} t/\hbar)$, and similarly for $| II \rangle$" is true. These energy eigenstates also constitute a basis, so any initial state, e.g. $|1 \rangle$, can be written as a linear combination of them: $|1 \rangle = \frac{1}{\sqrt{2}} (| I \rangle + | II \rangle$).

Thus to determine time evolution of the initial state one can:
1) expand the initial state in the basis of energy eigenstates;
2) time-evolve the expanded state using the fact that each energy eigenstate picks up a phase factor with the corresponding energy eigenvalue in the exponent;
3) go back to the original basis;
4) interpret coefficients in front of the original basis vectors as probability amplitudes of fidning the system at a later time in those states.

That's very clear! Thank you! So orthogonality is a sort of "instantaneous" property?

 

[physicsworks]

That's very clear! Thank you! So orthogonality is a sort of "instantaneous" property?

I wouldn’t use the word “instantaneous”. Orthogonality is the reflection of mutual exclusiveness.

 

[Roby002]

So...a base stase can become another base state online due to external factor such as magnetic fiele, tunneling and so on?