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Let's say I have eigenstates $|x\rangle$ associated with measurement of position. I know that the eigenstates corresponding to their respective eigenvalues form a basis, let's call it $A$. Now let's say I want to expand a generic state vector $|ϕ\rangle$ in that basis formed by eigenstates. Now, what does it mean when physicists say that:

"$ϕ(x)$ is the coordinate of the state $|ϕ\rangle$ in the basis $A$?"

Any explanation or reference to similar kind of questions would be helpful.

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    $\begingroup$ Which physicists? I.e. What books are you reading? (it will help provide explanations and examples that are more useful to you). $\endgroup$ – Emilio Pisanty Oct 21 '17 at 17:36
  • $\begingroup$ Hi Emilio, for example in this note, int.washington.edu/users/dbkaplan/324_01/DiracNotation.pdf, in page 4, in the sentence right below equation 19, there is this particular statement. $\endgroup$ – Patrick Oct 21 '17 at 17:38
  • $\begingroup$ I didn't understand what's your problem... which sentence of the document is unclear? $\endgroup$ – Erik Pillon Oct 21 '17 at 17:51
  • $\begingroup$ What does it mean by "ψ(x) are the coordinates of the our state |ψ> in the |x> basis"? $\endgroup$ – Patrick Oct 21 '17 at 17:53
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If you think of kets $\vert\psi\rangle$ as vectors, i.e. elements of a vector space in the sense that you can take linear combination of them, and you think of these vector as column vectors, then you can think of $\langle\psi\vert$ as a row vector so that $\langle\phi\vert\psi\rangle$ is a scalar product since it gives a number.

If we imagine that there were discretely many $x$'s labelled by $i$, i.e. if our positions came as $\{x_{i-1}, x_i,x_{i+1}\ldots\}$, you could associate the number $\psi(x_i)$ (i.e. the function $\psi$ evaluated at $x_i$) as the component of $\vert \psi\rangle$ on the $i$'th basis vector $\vert x_i\rangle$, with $\langle x_i\vert\psi\rangle=\psi(x_i)$. In this way, $\vert \psi\rangle$ and $\vert x_i\rangle$ are huge column vectors with components $$ \vert\psi\rangle = \left(\begin{array}{c} \vdots \\ \psi(x_{i-1})\\ \psi(x_i)\\ \psi(x_{i-1})\\ \vdots \end{array}\right)\, ,\qquad \vert x_i\rangle = \left(\begin{array}{c} \vdots \\ 0\\ 1\\ 0 \\ \vdots \end{array}\right) \begin{array}{c} \\ \leftarrow\hbox{position $i$}\\ \\ \end{array}\, ,\qquad \langle x_i\vert\psi\rangle = \psi(x_i) $$
Of course the position is really continuous, so the index $i$ is not needed but it's still convenient to this of $\psi(x)=\langle x\vert\psi\rangle$ as the value of the vector $\vert\psi\rangle$ on the basis vector $\vert x\rangle$.

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  • $\begingroup$ Hey, great explanation. Just one doubt which has lasted long. Wavefunction is the inner product of ket psi and basis vectors x. That is its an inner product so is a scalar. But function is vector, we know that. So can a scale of a larger space ( larger hibert space) be defined as a vector in another space( function space). $\endgroup$ – Shashaank Jul 10 at 14:36
  • $\begingroup$ @Shashaank the wavefunction is a vector in the Hilbert space of functions, but of course the actual value $\psi(x_i)$ is a number, the component of this vector along a particular basis state. $\endgroup$ – ZeroTheHero Jul 11 at 15:49
  • $\begingroup$ ok. I understand. You mean that $\psi(x) $ at each x is a scalar but the complete set of these numbers is a vector which is the function $\psi(x) $. Right !But if that is so what is the difference between ket psi and wavefunction psi. Both are vectors. Are they in different spaces. And is my argument on the inner products above correct. For two kets you define the inner product as $\langle\phi\mid\psi\rangle$ without any reference to basis. But in you first equation itself $\delta(x - x') and \psi(x) $ are themselves position basis representations of ket x and ket psi. $\endgroup$ – Shashaank Jul 11 at 17:18
  • $\begingroup$ you are then defining the inner product on these things ( functions) through that integral if both ket psi and wavefunction psi are vectors in the hilbert space then what is the difference between them..... There should be a difference between wavefunction psi and ket psi... If both are vectors what is the difference. Shouldn't the difference be that the ket psi is an abstract vector in an abstract vector space. The wavefunction ( the set of all the function values at all x and not a single value of x which shall be scalar) is vector in another vector space, the function space $\endgroup$ – Shashaank Jul 11 at 17:22
  • $\begingroup$ and you come a step down in that function space from that abstract vector space of ket psi in 2 operations the first being the projection on the x basis n then defining an inner product definition for the function ( the set of all values for x) $\endgroup$ – Shashaank Jul 11 at 17:25
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I think that what are you asking is not completely a mathematical problem but a physical trick...

Since I think that you got stuck around eq. (19) of the document you uploaded, I'll try to explain what's going on there.

The aim is expand our state $|\psi \rangle$ in terms of uor basis vector $|x\rangle$.

We can rewrite the state $|\psi >$ in the following way

$$|\psi\rangle = \hat{1}|\psi\rangle $$ At this point you only have multipied you state by an identity matrix. then using (18) that is nothing but a vector identity, you can rewrite the product as

$$\sum_k |x\rangle\langle x||\psi\rangle $$

but since in a generally Hilbert space the dimension is infinite, we are allowed to consider the integral instead of the sum and then we have

$$|\psi\rangle = \int |x\rangle\langle x||\psi\rangle dx $$

and defining (here's the trick!) $ \psi(x) = \langle x|\psi \rangle$ we finally get the equivalence $$|\psi\rangle = \int \psi(x)|x\rangle dx $$

This $\psi(x) $ is nothing but our familiar wavefunction. In the present language, $\psi(x) $ are the coordinates of the our state $|\psi\rangle$ in the $|x\rangle$ basis, since you're only projecting your function on the element of the basis!

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  • $\begingroup$ Hey, great explanation. Just one doubt which has lasted long. Wavefunction is the inner product of ket psi and basis vectors x. That is its an inner product so is a scalar. But function is vector, we know that. So can a scale of a larger space ( larger hibert space) be defined as a vector in another space( function space). $\endgroup$ – Shashaank Jul 10 at 14:38

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