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I'm studying for my quantum mechanics 3 exam and I really can't get my head round Dirac notation. My understanding so far is that $\lvert\psi\rangle$ is some mathematical object which doesn't really mean anything by itself (I've heard mentions of dual space but I don't think I should get into that). Then if you want to represent this object in a certain space (or basis??) you can do things like $\langle x\rvert\psi\rangle = \psi(x)$ which I sort of understand. But is $\langle x\rvert\psi\rangle$ an inner product like $\langle\phi\rvert\psi\rangle$ is or is it more complicated than that?

For the most part I understand the idea of projecting $\rvert\psi\rangle$ onto a space. Then we moved on to momentum operators and this is where I get confused. My lecturer has written things like $\hat p \rvert p\rangle = p\rvert p \rangle$ and similarly $\langle x \lvert \hat p \rvert p \rangle = p \langle x \lvert p \rangle = p\psi_p(x)$. Now, what is $\rvert p \rangle$ in relation to $\rvert \psi \rangle$? and what is $\psi_p(x)$?

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In the following, I will consider that $|x\rangle$ is a proper state of the Hilbert space of interest, although this is not mathematically rigorous.

In that case, yes, $\langle x|\psi\rangle$ is the projection of $|\psi\rangle$ onto the state $|x\rangle$ (the state fully localized at the position $x$), much in the same way that $\langle \phi|\psi\rangle$ is the projection of $\psi$ over the state $\phi$.

Here, $|\psi\rangle$ is any (normalized) state. But one can be interested in the eigenstates of the momentum operator $\hat p$. Let's call $k$ one eigenvalue of $\hat p$. The associated eigenstate can be labelled by the corresponding eigenvalue, so let's call it $|k\rangle$. Thus, $\hat p|k\rangle=k|k\rangle$. Then we can wonder what is the projection of this state $|k\rangle$ on a state $|x\rangle$, which gives us the amplitude probability of a particle in state $|k\rangle$ to be at position $x$. We can then call this overlap $\langle x|k\rangle=\psi_k(x)$, the wavefunction associated to the state $|k\rangle$. Then, simple manipulations show that $\langle x|\hat p|k\rangle=k\psi_k(x)$.

The only difference with the OP's notations (to help understanding what is each object in the above equations) is that one usually call $|p\rangle$ the eigenstate of $\hat p$ with eigenvalue $p$. In that case, the presence or absence of ket and hat tells us what is each object (a vector, an operator, a real number).

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  • $\begingroup$ Thank you for the great answer. So $\rvert k \rangle$ is a single momentum eigenstate of $\rvert \psi \rangle$? Once you start dealing with $\rvert k \rangle$ aren't you only working with one of the momentum eigenstates of $\rvert \psi \rangle$? And hence "throwing away" all the other eigenstates and so you're not dealing with the same wavefunction anymore? Just to clarify, $\psi_k(x)$ would be the x-projection of a single momentum eigenstate of $\rvert \psi \rangle$? $\endgroup$ – Horro Mar 31 '17 at 9:06
  • $\begingroup$ No! $|\psi\rangle$ is a state! So it cannot have an "eigenstate". $|psi\rangle$ is just a generic name for a vector in the Hilbert space. The same way that $f(x)$ is a generic name for a function. $\endgroup$ – Adam Mar 31 '17 at 9:08
  • $\begingroup$ Oh, so $\rvert k \rangle$ is just a state (like $\rvert \psi \rangle$) that is an eigenstate of momentum? So it is no different to calling it $\rvert \psi \rangle$ or $\rvert \phi \rangle$ but we use the letter k (or p) to signify that it is an eigenstate of momentum? $\endgroup$ – Horro Mar 31 '17 at 9:11
  • $\begingroup$ If you want, you can write $\langle x|k\rangle=f_k(x)$, the $k$ just labeling the fact that we are working with the projection of the state $|k\rangle$ and the $x$ the fact that we are looking onto the projection on the state $|x\rangle$. $\endgroup$ – Adam Mar 31 '17 at 9:11
  • $\begingroup$ Yes to your second comment. $\endgroup$ – Adam Mar 31 '17 at 9:11
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My understanding so far is that $|\psi\rangle$ is some mathematical object which doesn't really mean anything by itself

Of course it means something, it is a vector in a Hilbert space. Remember your Linear Algebra lectures: A vector $\vec v$ might be written as $(5,0)^T$ in one basis and as $(3,4)^T$ in another, but it is always the same vector $\vec v$. In Dirac notation, $|\psi\rangle$ plays the role of $\vec v$.

But is $\langle x\rvert\psi\rangle$ an inner product like $\langle\phi\rvert\psi\rangle$ is or is it more complicated than that?

It is an inner product. I'll continue with the example from above: Let ($\vec e_1$, $\vec e_2$) be the basis in which $\vec v$ has the representation $(3,4)^T$. That means that $\vec e_1 \cdot \vec v = 3$ and $\vec e_2 \cdot \vec v = 4$: We take the inner product of the vector $\vec v$ with a basis vector $\vec v_k$ to get the $k$-th component.

The difference in QM is that there are not only two basis vectors, but infinitely many. One choice of basis is the position basis, it contains infinitely many vectors $|x\rangle$ (one for every number $x$). To get the representation of $|\psi\rangle$ in this basis, we take the inner products $\langle x \vert \psi \rangle$. The resulting "components" is what we call the wave function $\psi(x)$.

[At this point of writing, the answer of Adam appeared. I think our answers complement each other very well like this, so I'll stop writing now. Read his answer for the parts of your question concerning the momentum operator specifically]

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