Consider a state $|\psi\rangle$. While discussing the Schroedinger equation, we say $$\hat{H}|\Psi(t)\rangle=i\hbar\frac{\partial}{\partial t}|\Psi(t)\rangle$$

We also define the hamiltonian operator as $$\hat{H}=\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)$$

But does this not imply that you're differentiating across "dimensions" of the vector $|\Psi(t)\rangle$? Is there some property of Hilbert spaces which allows this? When I think of a standard vector in $\text{R}^3$, it's clearly not possible to do such differentiation because there are only three dimensions and they are not 'continuous' in any way.

This might be similar to the following question: Applying an operator to a function vs. a (ket) vector But I'm interested in knowing what the whole theory behind $\frac{\partial}{\partial x}|\Psi(t)\rangle$ vs $\frac{\partial}{\partial x}\Psi(t)$ is. Is the former undefined? If yes, how is it indicated that we need to switch to using a function (rather than a vector) while applying a hamiltonian operator? If no, what is a rigorous argument that we can do this cross-dimention differetiation in a hilbert space?


In general the Hamiltonian operator is $$\hat{H}= \frac{\hat{p}^2}{2m}+V(\hat{x}) $$

This operator acts on vectors in the Hilbert space (kets)

The form you gave is how the Hamiltonian acts on the wavefunction that you get in the position representation of a certain vector. Consider some state $|\psi\rangle$, the wave function in position representation is defined as $\psi(x):=\langle x |\psi\rangle$ where $|x\rangle$ are the eigenstates of the position operator $\hat{x}$ with eigenvalue $x$. Hence

$$ \langle x|\hat{H}|\psi\rangle=\frac{1}{2m}\langle x|\hat{p}^2|\psi\rangle+V(x)\psi(x)$$

we just have to calculate

$$\langle x|\hat{p}|\psi\rangle=\int dp \langle x|p\rangle\langle p|\hat{p}|\psi\rangle =\int dp \,p\langle x|p\rangle\langle p|\psi\rangle=\int dp \,\frac{p}{\sqrt{2\pi\hbar}}e^{i \frac{xp}{\hbar}}\psi(p)=-i\hbar{\partial_x}\int dp \frac{1}{\sqrt{2\pi\hbar}}\,e^{i \frac{xp}{\hbar}}\psi(p)=-i\hbar\partial_x\int dp\langle x|p\rangle\langle p|\psi\rangle =-i\hbar\partial_x\psi(x)$$

Now, if we're talking about how the Hamiltonian acts on wave function, for short hand we write $\hat{H}\psi(x):=\langle x|\hat{H}|\psi\rangle$, hence in position representation

$$\hat{H}=-\frac{\hbar^2}{2m}\partial_{x}^2 +V(x) $$

the Hamiltonian you gave had a small error in it.

  • $\begingroup$ Thanks. It will take some time to go through and understand. I've fixed the weird hamiltonian in the question. $\endgroup$ – Dokadozi Jul 29 at 11:00
  • 2
    $\begingroup$ the important thing is that the Hamiltonian with the spatial derivative acts on the wave function $\psi(x)$, not on the ket $|\psi\rangle$, see also this question $\endgroup$ – user2723984 Jul 29 at 11:21

As a rule of thumb, every quantum mechanical state is described as a vector in Hilbert space—you seem to understand this. But the most important thing to understand here, is that Hilbert spaces are extremely flexible. Loosely speaking, every vector space with a well-defined inner product is a Hilbert space, provided that the inner product of a vector with itself is always well-defined. If that sentence doesn’t make sense to you, please tell and I’ll add some context. Knowing about abstract vector spaces beyond the Euclidean ones is extremely useful!

Now onto the misconception in your question. You write that we can define the Hamiltonian operator as $\frac{-i\hbar}{2m} \frac{\partial^2}{\partial x^2} + V(x)$ (don’t forget the mass factor!), but this is not completely true. The Hamiltonian operator is the operator that has all the eigenstates of energy. Again, if you’re not exactly sure what eigenstates mean in Quantum mechanics, don’t hesitate to ask.

So if we have a system in some eigenstate of the Hamiltonian operator $\vert 1 \rangle$, we can be assured that any measurement of energy returns the value $E$ such that $\bar{H} \vert 1 \rangle = E\vert 1 \rangle$.

Because we want to do calculations with the Hamiltonian, and discover these energy eigenstates, we give the Hamiltonian a representation: we write it down into a form that we can use along with our state vectors. One of such representations is $\frac{-i\hbar}{2m} \frac{\partial^2}{\partial x^2} + V(x)$, but it is not the only one. Which representation you use depends fully on which problem you’re facing.

To further illustrate this. I’ll now introduce two quantum mechanical problems, requiring very different Hilbert spaces, along with very different Hamiltonian operators.

First of all, consider the situation in which we face a quantum mechanical system with three energy levels. We are interested in the system’s energy. We can then label these three energy states by their kets: $\vert 1 \rangle$, $\vert 2 \rangle$ and $\vert 3 \rangle$. In this case, a representation of the vectors and operators in $\mathbb{R}^3$ is very useful. Operators on such vectors are 3x3 matrices. I’m on my phone now, so I can’t write out the matrices and column vectors, but I’ll describe them. The ith energy state can be represented by the column vector whose ith entry is 1, with all other entries 0. The Hamiltonian is then not represented by $\frac{-i\hbar}{2m} \frac{\partial^2}{\partial x^2} + V(x)$, but by a matrix with the energy values on the diagonal and with zeroes elsewhere.

Second of all, consider the infinite one-dimensional potential well. We want to know where the particle is likely to be. Since we’re interested in the particle’s position, we will not describe the system’s state as a linear combination of energy eigenstates as before. Instead, we will elect to represent the system’s state as a wave function—quadratically integrable functions form a Hilbert space too! (If you don’t understand the last bit, tell me and I’ll elaborate) Formally, the wavefunction representation is the position basis representation while the previous example is the energy basis representation.

Now whenever we represent the system’s state as a wavefunction, we need to change the representation of our Hamiltonian accordingly. In the position basis, the Hamiltonian’s representation is $\frac{-i\hbar}{2m} \frac{\partial^2}{\partial x^2} + V(x)$! Understanding why the Hamiltonian’s representation looks like this is a little harder than the first example. Fundamentally, the Hamiltonian for a particle is exactly equal to $\frac{\hat{p}^2}{2m} + V(x)$ and the momentum operator is chosen accordingly. But that goes beyond the scope of this answer.


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