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?

  • $\begingroup$ "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 :) $\endgroup$ Oct 2 at 22:50
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    $\begingroup$ 1. Your eq. (2) is for a free particle where there is no potential. Why do you think this should include a potential energy? 2. For the meaning of the $\langle x\rvert$ bras, see physics.stackexchange.com/q/201425/50583 and its linked questions $\endgroup$
    – ACuriousMind
    Oct 2 at 22:52
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    $\begingroup$ 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. $\endgroup$ Oct 2 at 22:53
  • $\begingroup$ @TobiasFünke I added some motivation and a reference. Thank you for helping make my post better. $\endgroup$
    – Alex Byard
    Oct 2 at 23:03
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    $\begingroup$ 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. $\endgroup$
    – joseph h
    Oct 3 at 0:54

2 Answers 2

  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.

  2. $|x\rangle$ 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.

  3. $\langle x |$ is not an operator. It is a dual vector.

  4. $|\psi\rangle$ 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").

  • $\begingroup$ 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. $\endgroup$
    – Alex Byard
    Oct 2 at 23:10
  • $\begingroup$ @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 $x$. $\endgroup$
    – Andrew
    Oct 2 at 23:13
  • $\begingroup$ Okay, and 3 means a bra is like a row vector, or like the functional side of an inner product? $\endgroup$
    – Alex Byard
    Oct 2 at 23:14
  • $\begingroup$ @AlexByard Exactly. $\endgroup$
    – Andrew
    Oct 2 at 23:14
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    $\begingroup$ @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 $\vec{v}^T$. $\endgroup$
    – Andrew
    Oct 2 at 23:18

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 𝜓(𝑥,𝑡)=⟨𝑥|𝜓,𝑡⟩ ?

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 |𝜓,𝑡⟩ 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 ⟨𝑥| , mathematically? Can you also explain the difference between 𝜓(𝑥,𝑡) and the 𝜓 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.

  • $\begingroup$ 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? $\endgroup$
    – Alex Byard
    Oct 2 at 23:36
  • $\begingroup$ 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 $c=\psi(x,t)$? $\endgroup$
    – Alex Byard
    Oct 2 at 23:38
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    $\begingroup$ 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 $\endgroup$ Oct 3 at 2:30
  • $\begingroup$ So the $\frac{1}{\sqrt{2}}$ are analogous to the $\psi(x,t)$? $\endgroup$
    – Alex Byard
    Oct 3 at 15:17

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