What happened to the Hamiltonian (and bra-ket confusion)

Question

Edit: This is my first question. If I don't conform to community guidelines please comment and I will happily update the question to meet the requests of the community I am asking for help.

I am trying to understand notation in the following exposition relating to the Schrödinger equation. I am a mathematician studying Liouville quantum gravity, so I want to understand this so I can further appreciate QFT in general from a physics POV. It isn't hard for me to port over mathematical intuition to novel notation or concepts, but I do get stuck with trying to understand what things mean. The information here is taken from Srendicki.

So in basic quantum mechanics the time evolution of the system is described by the Schrödinger equation:

$$i\hbar\frac{\partial}{\partial t}|\psi,t\rangle=H|\psi,t\rangle \quad .\tag 1 $$

I understand that the Hamiltonian operator is the sum of the kinetic and potential energy operators. In the position basis, this equation becomes

$$i\hbar\frac{\partial}{\partial t}\psi(x,t)=-\frac{\hbar^2}{2m}\nabla^2\psi(x,t) \quad \tag 2$$

where $$\psi(x,t)=\langle x|\psi,t\rangle.\tag 3$$

I can see how the kinetic energy operator is given by $-\frac{\hbar^2}{2m}\nabla^2$, as it is the same as $\frac{p\cdot p}{2m}$. What happened to the potential energy operator in this equation?

Next, can you explain what is happening with eq. (3)? From my reading, I guess that $|\psi,t\rangle$ represents the state of the system, i.e., is a vector of quantities that can tell us everything we want to know about the system. I also take it that bras are operators. What operator is $\langle x|$, mathematically? Can you also explain the difference between $\psi(x,t)$ and the $\psi$ in the ket?

"What happened to the potential energy operator in this equation" what do you mean exactly? You are just stating some things, without any motivation or reference. The question does not make sense. I've edited the math, consider to have a look for possible future posts :) — Tobias Fünke, Oct 02 '23 at 22:50
Briefly: Yes, if the Hamiltonian is the sum of the kinetic energy + potential energy operator, then in the position representation there clearly should be the position representation of the potential energy operator, which often is written as $V(x)$ in your case. Regarding the bra-ket/Dirac notation: Use the search function, there are plenty questions here dealing with this issue. If you then still have a specific question, consider to edit the question. — Tobias Fünke, Oct 02 '23 at 22:53
@TobiasFünke I added some motivation and a reference. Thank you for helping make my post better. — Alex Byard, Oct 02 '23 at 23:03
@ACuriousMind I didn't recognize that this was a system without potential. Srendicki does say that this Hamiltonian is for a spinless, nonrelativistic particle with no forces acting on it, which I missed. How does the spin, relativistic considerations, and addition of forces change the Hamiltonian? I.e, how would I know what the Hamiltonian is for a general system? — Alex Byard, Oct 02 '23 at 23:07
"no forces acting on it" = "no potential", this is a basic idea of classical physics. Sounds like you're trying to run before you can walk: Almost all expositions of quantum mechanics presuppose some familiarity with classical mechanics (and in particular the more you know about classical Hamiltonian mechanics the less mysterious the quantum version of it will be). — ACuriousMind, Oct 02 '23 at 23:54
@ACuriousMind This was a "damn I didn't read that sentence" mistake not a life-shattering "I don't actually know any physics" mistake. You're part right, though. I'm a mathematician, and it seems that in my line of work I'll forever be stuck solving problems interesting to physicists for some appreciable but inaccessible reason. Hence, the goal of this post. My recent research has been on the DOZZ formula, which makes total sense mathematically, though I can't pretend I know what it really means to "solve" a CFT beyond "computing the correlation functions", if you catch my drift. — Alex Byard, Oct 03 '23 at 00:02
@ACuriousMind Perhaps I'm abusing the "I know a lot of math so all I have to learn is the physics" point of view a bit too much... For instance, the past couple weeks have made me terrified of natural units since I'm certain I'll reach a point where I want factors of the speed of light and they'll just be gone for good (modulo dimensional analysis I don't know how to do...) — Alex Byard, Oct 03 '23 at 00:03
How does the spin, relativistic considerations, and addition of forces change the Hamiltonian? That is another question altogether, and the answer would need to include a lot of information. So much so that the question will probably be closed for being unfocused. Consider asking a new question, but be a lot more specific. — joseph h, Oct 03 '23 at 00:16
@josephh I unfortunately feel like I don't know enough to ask a more specific question modulo just picking one of the three, e.g., how does just spin affect the Hamiltonian? I fear this will still be interpreted as being too broad. This means that I need to learn more before asking questions, but I don't know where to learn more. Do you have any references that can help me narrow my question down? — Alex Byard, Oct 03 '23 at 00:22
Perhaps the most notable thing that changes when you include spin is that in general, the states $\psi$ become spinors (vector like objects with at least 2 components) and the classical Hamiltonian you have in your OP is different. The treatment in such cases is such that one adopts relativistic quantum mechanics and perhaps quantum field theory (unless you can use the nonrelativistic Pauli equation). Questions like that have been asked on this site before, and so please use the search function so that your question is not closed as a duplicate. Like I said earlier, be specific. Cheers. — joseph h, Oct 03 '23 at 00:54
@Qmechanic This might have made this easier to understand/answer... — Alex Byard, Oct 03 '23 at 15:17

score 1 · Answer 1 · answered Oct 02 '23 at 23:00

1

In general, the Hamiltonian operator $\hat H$ is "some function" of the momentum $\hat p$ and position $\hat x$ operators (which do not commute in quantum mechanics, so ordering matters). In your equation 2, you've chosen a specific form of $\hat H$, which is valid, but not general. If you want $\hat H$ to have the form of kinetic energy plus potential energy, you should take $$ \hat H = \frac{\hat p^2}{2m} + V(\hat x) $$ In the position basis, $\hat H$ becomes a differential operator which acts on the wavefunction $\psi(x)$ like $$ \hat H \psi(x) = -\frac{\hbar^2}{2m}\nabla^2 \psi(x) + V(x) \psi(x) $$ However, in general, you can imagine more complicated forms of the Hamiltonian.
$|x\rangle$ is a state or a vector in Hilbert space. It is defined by the eigenvalue equation $$ \hat{x} | x\rangle = x | x \rangle $$ where on the left hand side $\hat{x}$ is an operator, and on the right hand side $x$ is a real number. The $x$ appearing inside the bracket $|x\rangle$ is just a label for this state.
$\langle x |$ is not an operator. It is a dual vector.
$|\psi\rangle$ is a basis-independent way to refer to the state $\psi$. Mathematically, $|\psi \rangle$ is a vector in Hilbert space (or a ray if it represents the entire state of a system, which is only defined up to an overall phase factor). $\psi(x) = \langle x | \psi \rangle$ are the coefficients of the state $|\psi\rangle$ in the position basis. Mathematically, $\psi(x)$ is a complex-valued function of $x$ (the "wavefunction").

answered Oct 02 '23 at 23:00

Andrew

48,573

From 2. I'm taking away that a state is a line in a Hilbert space, and that it is an eigenvalue of some (?) operator. – Alex Byard Oct 02 '23 at 23:10
@AlexByard A general state is a vector in Hilbert space. The state $|x\rangle$ is an eigenvector of the position operator $\hat{x}$ with eigenvalue $x$. – Andrew Oct 02 '23 at 23:13
Okay, and 3 means a bra is like a row vector, or like the functional side of an inner product? – Alex Byard Oct 02 '23 at 23:14
@AlexByard Exactly. – Andrew Oct 02 '23 at 23:14
And then a state $|\psi\rangle$ is like a way to represent a basis independent quantity, and the dual vector $\langle x|$ throws us in that basis? Or is it a bit different than that – Alex Byard Oct 02 '23 at 23:15
I'm thinking like how we talk about representing tensors by their transformation laws. A basis independent "state" which can be measured off a basis? – Alex Byard Oct 02 '23 at 23:17
1

@AlexByard Basically. To make an analogy finite dimensional vector spaces, imagine $\vec{v}$ is a generic vector, while $\hat{e}_x$ is a unit basis vector in the $x$ direction. Then $v(x)=\vec{x}\cdot\hat{e}_x$ is the $x$ component of $\vec{v}$. $|\psi\rangle$ is like $\vec{v}$, $|x\rangle$ is like $\hat{e}_x$, and $\psi(x)$ is like $v(x)$. Also, $\langle x |$ is like $\hat{e}_x^T$, and $\langle \psi|$ is like $\vec{v}^T$. – Andrew Oct 02 '23 at 23:18
This makes me a bit confused. I am used to changing bases, but not necessarily introducing them. If $|\psi\rangle$ is a basis-independent quantity describing the state corresponding to the label $\psi$, then how does it make sense to think of $\langle x|$ as $e_x^\top$? I feel like treating it as an inner product actually requires that we have concrete entries for $|\psi\rangle$ in some basis. – Alex Byard Oct 02 '23 at 23:25
@AlexByard I'm not sure I understand your question but I recommend translating it into the finite dimensional analogy, since (at a physicist level of rigor) the answers tend to be the same. You don't need to introduce a basis to define a vector $v$ abstractly, just like you don't need to introduce a basis to define a state $|\psi\rangle$ abstractly. If you want concrete values, then for a vector you need to take an inner product with a basis vector, like $e_x^T v=v_x$, and for a state you need to take an inner product like $\langle x | \psi \rangle = \psi(x)$. – Andrew Oct 02 '23 at 23:32
So $|\psi\rangle$ represents an element (or line) of your Hilbert space. The dual vector $\langle x|$ doesn't care what its entries relative to a basis are, just as long as it comes from the Hilbert space. As a dual vector $\langle x|$ maps us from our Hilbert space to $\mathbb{C}$, and gives the probability amplitude $\psi(x,t)$ of the state. Is this correct? – Alex Byard Oct 02 '23 at 23:45
Also, is $\langle x|$ an explicit dual vector? Like $\psi((1,0,0),t)=\langle1,0,0|\psi\rangle$? – Alex Byard Oct 02 '23 at 23:47
I was also getting at the fact that $e_x^\top v=v_x$ is dependent on the calculational framework of linear algebra. We could explicitly calculate $(1,0,0)(v_x,v_y,v_z)^\top=v_x$, but this requires us to have a $v_x$ to extract. I don't know how to do this without knowing $v_x,v_y,v_z$ in some basis first. – Alex Byard Oct 02 '23 at 23:50

Relativisticcucumber · Answer 2 · 2023-10-03T02:28:54.673

What happened to the potential energy operator in this equation?

The potential is still there. The Hamiltonian in position basis for the time dependent Schrodinger equation is $$H = \frac{\hbar^2}{2m}\nabla^2 + V(x,t)$$

Next, can you explain what is happening with (,)=⟨|,⟩ ?

In quantum mechanics, we have states that are independent of the basis they are expressed in. $\vert \psi \rangle$ is a state (a ket). Now, we can use the projection operator to project this state into a basis $$\vert x \rangle \langle x \vert \psi \rangle = c\vert x \rangle$$ where $c$ is the probability amplitude of the state being in the projected state and can be expressed as $c = \langle x \vert \psi \rangle$. This state is written, perhaps in bad notation, as $\psi(x)$.

From my reading, I guess that |,⟩ represents the state of the system, i.e., is a vector of quantities that can tell us everything we want to know about the system. I also take it that bras are operators. What operator is ⟨| , mathematically? Can you also explain the difference between (,) and the in the ket?

To emphasize oncemore: $\vert \psi \rangle$ is basis independent, and $\psi (x,t)$ is written in the position basis. $\langle x \vert$ is a dual vector, not an operator. :)

Addendum: If we have a state, it can be written as a linear combination of it’s basis vectors. It is a postulate of quantum mechanics that physical (i.e. allowed) states must be normalized due to the born interpretation. Thus, what we are actually doing when we apply the projection operators is inserting a complete set of eigenstates $\vert a \rangle$. $$\sum_a \vert a \rangle \langle a \vert$$ Due to the orthonormality of states, the projection operators will only act on the states orthonormal to them, and the rest will vanish. This is easier to see with a 2-state system. Say our basis is $\vert a_1 \rangle = \vert \uparrow \rangle$ and $\vert a_2 \rangle = \vert \downarrow \rangle$. Now, let’s define a state $\vert \psi \rangle = \frac{1}{\sqrt{2}}\vert \uparrow \rangle + \frac{1}{\sqrt{2}}\vert \downarrow \rangle$ Then, when we insert the completeness relation, we have $$\vert \psi \rangle = \sum_{a} \vert a \rangle \langle a \vert \psi \rangle = (\vert \uparrow \rangle \langle \uparrow \vert + \vert \downarrow \rangle \langle \downarrow \vert)(\frac{1}{\sqrt{2}}\vert \uparrow \rangle + \frac{1}{\sqrt{2}}\vert \downarrow \rangle)$$ Simplifying this using orthonormality, we get $\vert \psi \rangle = \frac{1}{\sqrt{2}}\vert \uparrow \rangle + \frac{1}{\sqrt{2}}\vert \downarrow \rangle$, our original state. This makes sense because the state was already in the spin 1/2 basis. Now, if we square the coefficients of each term, both are $\frac{1}{2}$, which sums up to 1. We interpret this as the state is in a superposition with a probability of $\frac{1}{2}$ of finding the position in either state $\vert \uparrow \rangle$ or $\vert \downarrow \rangle$ upon measurement. That is why we call these coefficients probability amplitudes. The position basis is a continuous version of this argument.

Oh, maybe this makes more sense than I thought. For a vector space, a dual vector is a map $V\to\mathbb{C}$, so then $|\psi\rangle$ doesn't need to have particular entries as long as it's alright with representing a vector (line) itself (so far as it's an element of $V$). Does this mean that the dual vector $\langle x|$ doesn't have particular entries as well? Also, is $|\psi\rangle$ implicitly time/position-dependent, so that writing the amplitude as $\psi(x,t)$ makes sense? — Alex Byard, Oct 02 '23 at 23:36
When you say that $c$ is the probability of the state being in the projected state, what does this mean? Really, $c$ must represent the same amplitude independent of basis, right? Do we also call the probability amplitude the state, if $c=\psi(x,t)$? — Alex Byard, Oct 02 '23 at 23:38
I wrote an addendum to address your second question. Does it clarify? For the first question, I'm not exactly sure what you mean. @AlexByard — Relativisticcucumber, Oct 03 '23 at 02:30
So the $\frac{1}{\sqrt{2}}$ are analogous to the $\psi(x,t)$? — Alex Byard, Oct 03 '23 at 15:17

What happened to the Hamiltonian (and bra-ket confusion)

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