# Hamiltonian Operator

I've learned that the Hamiltonian Operator corresponds to the total energy of the system when applied to a general wave function. After applying and obtaining the measurement (energy), the wave function turns into the eigenfunction.

Does the wave function before the operator is applied have to be an eigenfunction of the operator, or can the operator be applied to any wave function?

• Please do not change your question after it has been answered; it is intentional that you cannot delete questions that have received upvoted answers. Commented Oct 22, 2023 at 11:00

It is not correct that applying the Hamiltonian gives you the energy of the system if the system is in a general state, $$|\psi\rangle$$.

Upon measurement (in an experiment for example) the system "jumps" from whatever state it was in before (say $$|\psi\rangle$$) to some eigenstate of the Hamiltonian, which, as you say is the operator that corresponds to energy in quantum mechanics:

$$|\psi\rangle \xrightarrow{\mathrm{Measurement }\space\mathrm{of}\space\mathrm{Energy}}|E_i\rangle.$$

Note that "applying the Hamiltonian" to $$|E_i\rangle$$ will return the same state multiplied by the eigenvalue (as here we have an eigenvector of $$H$$, not just any old state):

$$H|E_i\rangle=E_i|E_i\rangle.$$

Again, just to really emphasise, measurement of the energy of the system is what causes the system to "turn into" an eigenfunction of the Hamiltonian, not applying the Hamiltonian.

I assume you are working in the context of textbook non-relativistic quantum mechanics.

Recall that states of a quantum system are vectors$$^{1}$$ in a complex vector space (more pedantically, Hilbert space). That is, a state $$\lvert \psi \rangle$$ lives in the complex vector space $$\mathcal{H}$$. More concisely, we say that $$\lvert \psi \rangle \in \mathcal{H}$$.

Any observable, including the Hamiltonian, may be formally described as a linear map over the space of states $$\mathcal{H}$$. This is written in mathematical language as $$H: \mathcal{H} \rightarrow \mathcal{H}$$. Hence, the action of $$H$$ on any $$\lvert \psi \rangle \in \mathcal{H}$$ is defined.

As Charlie answers, it is important to not conflate measuring the energy of a system with applying the Hamiltonian operator. I.e., applying the Hamiltonian operator is not what we mean by measuring the energy of a system.

Rather, it is true that after measuring a system's energy, it is postulated to be found in an energy eigenstate $$\lvert E_n \rangle$$. The definition of an energy eigenstate is that the relation $$H\lvert E_n \rangle = E_n \lvert E_n \rangle$$ holds, where $$E_n$$ is some scalar value.

1. More pedantically, a state $$\lvert \psi \rangle$$ of a quantum system is a representative with unit norm of a ray in Hilbert space.
• ...and even more pedantically, a state should be normalized to unity. And even more, although this is far beyond the scope of this answer, the Hamiltonian is in general unbounded and defined only on a dense proper subspace. Commented Oct 21, 2023 at 23:18
• @TobiasFünke I tried to edit to the best of my ability and level of my knowledge :) Commented Oct 21, 2023 at 23:24

I started to make this a comment because it is so short, but it really is an answer so:

The operator can be applied to any wave function.

The process of measurement, leads to a collapse of the wave function to one of the eigenstates with a certain probability. The eigen basis which gives this probability distribution is definitely related the measurement apparatus.

For example, consider a system described by a wave function $$|\Psi\rangle \in \mathcal{H}$$, where $$\mathcal{H}$$ is complex vector space or Hilbert space. Now, according to postulates of quantum mechanics every observable has a corresponding operator defined on $$\mathcal{H}$$. Let's consider on such operator that corresponds to the total energy, which happens to be the Hamiltonian operator $$\hat{\mathbb{H}}$$.

Here we define $$\hat{\mathbb{H}}:\mathcal{H}\rightarrow\mathcal{H}$$ as a linear transformation that maps any $$|\Phi\rangle \rightarrow |\Theta\rangle$$ where $$|\Phi\rangle,|\Theta\rangle \in \mathcal{H}$$.

Now, when we say we make a measurement, let's assume that it would be associated with an observable. Then, there should be an operator on $$\mathcal{H}$$ that is associated to such a measurement. Mathematically, we can define a measurement apparatus using the eigen states of the operator associated with the observable we want to measure. So, a measurement apparatus that measures the energy can be defined as, $$\hat{\mathbb{P}}_{\varepsilon} = \sum_{i} |\chi_{i}\rangle\langle\chi_{i}| = \mathbb{1}$$ where $$|\chi_{i}\rangle$$'s are eigen kets associated with $$\hat{\mathbb{H}}$$ where $$\hat{\mathbb{H}}|\chi_{i}\rangle=\varepsilon_{i}|\chi_{i}\rangle$$. Now, projecting this operator into its own eigen basis gives us, $$\hat{\mathbb{P}}_{\varepsilon}\hat{\mathbb{H}} = \sum_{i} |\chi_{i}\rangle\langle\chi_{i}| \hat{\mathbb{H}}\\\hat{\mathbb{H}} = \sum_{i} |\chi_{i}\rangle\varepsilon_{i}\langle\chi_{i}|$$

Now, if you have the Hamiltonian act on the wave function we get,

$$\hat{\mathbb{H}} |\Psi\rangle = \sum_{i} |\chi_{i}\rangle\varepsilon_{i}\langle\chi_{i} |\Psi\rangle\\ = \sum_{i} \varepsilon_{i} c_{i}|\chi_{i}\rangle$$

where $$c_{i} = \langle\chi_{i} |\Psi\rangle$$ is the amplitude of the $$i$$th eigen state, or in other words, $$|c_i|^2$$ is probability for the wave function $$|\Psi\rangle$$ to collapse to the $$i$$th energy eigen state upon measurement.

Further, it also follows that the expectation value of the Hamiltonian is given as,

$$E_{\Psi} = \langle \Psi |\hat{\mathbb{H}}|\Psi\rangle = \sum_{i} \varepsilon_{i} |c_{i}|^{2}$$

To summarize, the collapse of the wave function $$|\Psi\rangle$$ upon measurement, is dependent on the measurement apparatus i.e. it collapses to one of the eigen states of the observable being measured with a certain probability. Finally, when you make measure in quantum mechanics, what you measure are the probability distributions.