Suppose, we have a basis $|u\rangle$, described by the function $u=g(x)$. We can normalize this basis, using our standard $|x\rangle$ basis using the following :

$$\hat{I}=\int |x\rangle\langle x|dx=\int |x\rangle\langle x|\frac{du}{g'og^{-1}(u)}=\int |x\rangle\langle x|h(u)du=\int |u\rangle\langle u|du$$

Now, we can set $|u\rangle = \sqrt{h(u)}|x\rangle$ and we have managed to normalize it. Moreover, since $u=g(x)$, we can write $|u\rangle=|g(x)\rangle$ and $|x\rangle=|g^{-1}(u)\rangle$.

Suppose we have an abstract wavefunction given by $|\psi\rangle$. Let us assume this is normalized.

In the $x$, and the $u$ basis, this can be shown as :

$$\langle\psi|\psi\rangle=\int |\psi(x)|^2dx=\int |\Psi(u)|^2dx=1$$

Here $\psi(x)=\langle x|\psi\rangle$ and $\Psi(u)=\langle u|\psi\rangle$.

However, we know that $\psi(x)=\psi(g^{-1}(u))=\phi(u)$.

We can easily check that :

$$\int |\phi(u)|^2du \ne 1$$

In fact, we can show using simple substitution, that :

$$\langle\psi|\psi\rangle=\int |\phi(u)|^2h(u)du=1$$

My question is, does this $\phi(u)$ represent a new abstract wave-function since it is not normalized in the integral form. Or is it nothing but the representation of the original ket $|\psi\rangle$ in the $|g^{-1}(u)\rangle$ basis?

So, should we write $\phi(u)=\langle g^{-1}(u)|\psi\rangle$ ? Or should we write $\phi(u)=\langle u|\phi\rangle ? $

What I've done is, I've taken $\langle g^{-1}(u)|\psi\rangle$ and I've absorbed the $g^{-1}$ function into the ket, to create a new ket $|\phi\rangle$.

Hence, we have :$$\langle g^{-1}(u)|\psi\rangle = \langle u|\phi\rangle$$

Is this, strictly speaking, allowed ? Or should we interpret $\phi(u)=\psi(x)$ purely as, either $\langle g^{-1}(u)|\psi\rangle$ or $\langle u|\phi\rangle$ ?

What I think is, since $\phi(u)$ is not directly normalized in the integral form, it should represent a different abstract wavefunction. However, that would suggest, that as I did a simple $u$ substitution, replace all the $x$'s in the integral by $g^{-1}(u)$'s, and taken the appropriate jacobian, I've also changed the abstract wavefunction.

While mathematically this doesn't make any difference whatsoever, I'd like to know, if this is allowed in the physical sense.


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    $\begingroup$ I am confused. What do you mean when you say that you have a basis $|u\rangle$ given by $u=g(x)$? For example, the position basis is a collection of (generalized) kets which we label via their eigenvalue when acted upon by $\hat X$. Are you saying that $\langle x|u\rangle = g(x)$? Or are you simply relabeling the $|x\rangle$ basis $|x\rangle \mapsto \widetilde{|u\rangle}$ such that $|1\rangle,|2\rangle,|3\rangle \mapsto |\widetilde{g(1)}\rangle,|\widetilde{g(2)}\rangle,|\widetilde{g(3)}\rangle \ldots$ so $\hat X\widetilde{|u\rangle}=g^{-1}(u)\widetilde{|u\rangle}$? $\endgroup$
    – J. Murray
    Commented Nov 30, 2021 at 20:59
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    $\begingroup$ @J.Murray This is a question in a long sequence of questions about these kinds of things. It's worth looking at the questions and answers I linked below, along with some of other similar questions that the OP has asked. (Might also be worth looking at my particular interpretation of what's going on here in my answer below.) $\endgroup$
    – march
    Commented Nov 30, 2021 at 21:05

1 Answer 1


Part 1: what can we "absorb" into kets?

I think that some of the confusion arises because multiplying by $\frac{1}{\sqrt{g'(x)}}$ means different things when performed on an abstract state versus a position-space (or $u$-space) wave function. Let's consider the function $u=g(x)$ and the corresponding operator $\hat{u}=g(\hat{x})$ with eigenvalues $u=g(x)$. For convenience, let's suppose that $g$ is monotone increasing and onto.

The most important observation to make is that the product $$ \frac{1}{\sqrt{g'(x)}}\left\lvert \psi \right\rangle $$ doesn't change the state. Instead, this is just multiplication of a ket vector by a number because in this representation, $x$ is a number! It happens to be an eigenvalue of the operator $\hat{x}$, but in the product above, that doesn't really matter. However, when we do $$ \frac{1}{\sqrt{g'(x)}}\psi(x)\,, $$ the only real way to interpret this is as the action of the multiplication operator $\left(\sqrt{g'(\hat{x})}\right)^{-1}$ on the state $\left\lvert \psi \right\rangle$, represented in the expression above as $\psi(x)$.

The difference between these two occurs because in the second, there is a hidden $\left\langle x \right\rvert$, whereas in the first, there is not. That is, defining the new state $$ \left\lvert \phi \right\rangle = \frac{1}{\sqrt{g'(\hat{x})}} \left\lvert\psi \right\rangle\,, $$ we have that $$ \frac{1}{\sqrt{g'(x)}}\psi(x) = \frac{1}{\sqrt{g'(x)}}\left\langle x | \psi \right\rangle = \left( \left\langle x \right\rvert \frac{1}{\sqrt{g'(x)}}\right)\left\lvert \psi \right\rangle = \left(\left\langle x \right\rvert \frac{1}{\sqrt{g'(\hat{x})}}\right)\left\lvert \psi \right\rangle = \left\langle x \right\rvert \left( \frac{1}{\sqrt{g'(\hat{x})}}\left\lvert \psi \right\rangle\right) = \left\langle x | \phi \right\rangle = \phi(x)\,, $$ but we can't actually do anything with $$ \frac{1}{\sqrt{g'(x)}}\left\lvert \psi \right\rangle=\,? $$ because the state $\left\lvert \psi \right\rangle$ is not (necessarily) an eigenstate of $\hat{x}$, and so we can't think of this as resulting from the action of $\left(\sqrt{g'(\hat{x})}\right)^{-1}$ on $\left\lvert \psi \right\rangle$.

Thus, it doesn't make sense to absorb a factor of $\frac{1}{\sqrt{g'(x)}}$ into an abstract ket vector $\left\lvert \psi \right\rangle$ in order to define a new state. It doesn't make a new state!$-$because it's just the multiplication of $\left\lvert \psi \right\rangle$ by the number $\frac{1}{\sqrt{g'(x)}}$. To understand what the transformation $x=g^{-1}(u)$ is actually doing requires more careful calculations (see below under "Some extra context").

Part 2: What is the context of $\langle g^{-1}(u)|\psi\rangle = \langle u|\phi\rangle$?

By setting $$ \langle x|\psi\rangle=\langle g^{-1}(u)|\psi\rangle = \langle u|\phi\rangle\,, $$ we are asking the following question. What quantum state (up to normalization), when represented in $u$-space has the same wave function as $\left\lvert \psi \right\rangle$ represented in $x$-space, with $u=g(x)$? Let's do the same calculations as above: $$ \langle u|\phi\rangle = \frac{\left\langle g^{-1}(u) \right\rvert }{\sqrt{g'(g^{-1}(u))}}\left\lvert \phi \right\rangle = \left\langle x \right\rvert\frac{1}{\sqrt{g'(x)}}\left\lvert \phi \right\rangle = \left\langle x \right\rvert\frac{1}{\sqrt{g'(\hat{x})}}\left\lvert \phi \right\rangle =\langle x | \psi\rangle\,, $$ where $$ \left\lvert \psi \right\rangle = \frac{1}{\sqrt{g'(\hat{x})}} \left\lvert\phi \right\rangle\,,. $$ This transformation is the opposite of the one that we consider before, that was (in a loose sense) equivalent to the change of basis from $\left\lvert x \right\rangle$ to $\left\lvert u \right\rangle$. Thus, the answer to "is this allowed" is yes, and what you are actually doing to construct the new state is applying the inverse of $\frac{1}{\sqrt{g'(\hat{x})}}$ to the original state $\left\lvert \psi \right\rangle$.

However, it's unclear what physical context this might come up in.

Some extra context

To really be careful about these kinds of transformations, it pays to work in Dirac notation, because it all depends on what you want to do with it, and what you're starting from. The calculations above don't uniquely determine whether the state you've constructed is a new state or a new representation of the same state, the reason being that you haven't actually brought in any of the interpretational apparatus of quantum mechanics yet. That all comes in via expectation values of operators and the Schrodinger equation. So let's see if I can disentangle this a little bit.

New representations of the same state

This viewpoint is taken in this answer to the previous question Change of Basis in quantum mechanics using Bra-Ket notation. The idea there is to bring in the probabilistic interpretation of the wave function from the beginning by defining the projection operator $$ \hat{P}_{u,[a,b]} = \operatorname{Pr}(a\leq u\leq b)\,, $$ as a "piece" of the resolution of the identity. That is, given the transformation $u=g(x)$ (assumed to be monotonic increasing and onto, for convenience), the resolution of the identity is given by $$ \hat{I} = \int_{-\infty}^{\infty} dx\,\left\lvert x \right\rangle\left\langle x \right\rvert = \int_{-\infty}^{\infty}\frac{du}{g'(g^{-1}(u))} \left\lvert g^{-1}(u) \right\rangle \left\langle g^{-1}(u) \right\rvert = \int_{-\infty}^{\infty}du\, \left\lvert u \right\rangle \left\langle u \right\rvert\,, $$ where, as in the OP, we have defined $\left\lvert u \right\rangle = \left\lvert g^{-1}(u) \right\rangle/\sqrt{g'(g^{-1}(u))}$; note that $\left\lvert x \right\rangle = \left\lvert g^{-1}(u) \right\rangle$. Finally, then, the probability of measuring the operator $\hat{u} = g(\hat{x})$ when in the state $\left\lvert \psi \right\rangle$ and getting a value in $[a,b]$ is given by $$ \left\langle \psi \right\rvert\hat{P}_{u,[a,b]}\left\lvert \psi \right\rangle = \int_a^bdu\, \left\langle \psi | u \right\rangle \left\langle u | \psi \right\rangle = \int_a^bdu\,\left\lvert \left\langle u | \psi \right\rangle\right\rvert^2 \,, $$ in which case we can pick out the $u$-representation of the state $\psi$ as $\psi_{\hat{u}}(u)= \left\langle u | \psi \right\rangle$. The standard $x$-representation of the state is just $\psi_{\hat{x}}(x)= \left\langle x | \psi \right\rangle$.$^1$

$^1$An aside on notation is warranted. Since we are always considering the same abstract state $\left\lvert \psi \right\rangle$, I want to denote the $x$- and $u$-space wave functions with the same symbol, but I also want to notate the fact that these two functions have different functional forms. Hence, I've gone with the subscript of the name of the operator. There's no universal way of doing this. Sometimes we use tildes, but I tend to restrict the use of tildes for Fourier transforms, i.e., momentum-space representations.

A new state

Functionally, the two different representations are related as follows: $$ \psi_{\hat{u}}(u)= \left\langle u | \psi \right\rangle = \frac{\left\langle g^{-1}(u) \right\rvert }{\sqrt{g'(g^{-1}(u))}} \left\lvert\psi \right\rangle =\frac{1}{\sqrt{g'(x)}}\left\langle x | \psi \right\rangle =\frac{1}{\sqrt{g'(x)}}\psi_{\hat{x}}(x) =\frac{1}{\sqrt{g'(g^{-1}(u))}}\psi_{\hat{x}}(g^{-1}(u)) \,. $$ The two different functional forms $\psi_{\hat{x}}$ and $\psi_{\hat{u}}$ are related mathematically in this way, but if we use the interpretation in the previous section, we are using these two different functional forms to represent the same state in different (in a sense) bases.

However, there's a different way to interpret the manipulation that we've just done. Instead of pulling the square root of the measure out from the inner product, we can instead interpret this number as the action of a Hermitian operator from the right on the basis state, which is what's done in this answer to the question Dirac notation, integral and change of basis. That is, starting from the third entry in the previous derivation, we have $$ \frac{\left\langle g^{-1}(u) \right\rvert }{\sqrt{g'(g^{-1}(u))}} \left\lvert\psi \right\rangle = \left(\left\langle x \right\rvert \frac{1}{\sqrt{g'(x)}} \right)\left\lvert\psi \right\rangle = \left(\left\langle x \right\rvert \frac{1}{\sqrt{g'(\hat{x})}} \right)\left\lvert\psi \right\rangle = \left\langle x \right\rvert \left(\frac{1}{\sqrt{g'(\hat{x})}} \left\lvert\psi \right\rangle\right) \,. $$ (Notice the passage from $x$ to $\hat{x}$!) Then, we have created a new abstract state $\left\lvert \phi \right\rangle$, given by $$ \left\lvert \phi \right\rangle = \frac{1}{\sqrt{g'(\hat{x})}} \left\lvert\psi \right\rangle\,, $$ by acting on our original state $\left\lvert \psi \right\rangle$ with some operator. This state is of course not normalized (this is directly related to observations in the OP$^2$), as rarely is it the case that when we make a new state by operating with some operator do we get a state that is normalized (in fact, to make this generally true, the operator would have to be unitary).

Now, what are the $x$- and $u$-space representations of this new state? We can just follow the standard prescription, given by $$ \phi_{\hat{x}}(x) = \left\langle x | \phi \right\rangle = \left\langle x \right\rvert \frac{1}{\sqrt{g'(\hat{x})}} \left\lvert\psi \right\rangle = \left\langle x \right\rvert \frac{1}{\sqrt{g'({x})}} \left\lvert\psi \right\rangle = \frac{1}{\sqrt{g'({x})}} \left\langle x | \psi \right\rangle = \frac{1}{\sqrt{g'({x})}}\psi_{\hat{x}}(x)\,, $$ and, similarly, $$ \phi_{\hat{u}}(u) = \left\langle u | \phi \right\rangle = \frac{1}{\sqrt{g'(x)}}\phi_{\hat{x}}(x) =\frac{1}{\sqrt{g'(x)}} \frac{1}{\sqrt{g'({x})}}\psi_{\hat{x}}(x)\,. $$

$^2$Although it's hard to figure out what I should call $\phi$, $\Psi$, and $\psi$ to match the OP's notation. This is why I have tried to be as unambiguous as possible in notating the wave function, and it's why I try to work in the Dirac language as much as possible.

A new representation or a new state?

We can finally make the connection based on the following observation. Note that we have derived, in different ways, the functional relationships $$ \psi_{\hat{u}}(u) = \frac{1}{\sqrt{g'({x})}}\psi_{\hat{x}}(x) = \phi_{\hat{x}}(x) = \sqrt{g'(g^{-1}(u))} \phi_{\hat{u}}(u) \,. $$ Note, however, that these functional relationships occur because we are using the same operator $\hat{u} = g(\hat{x})$ but in different contexts. In one case$-$the representations of $\left\lvert \psi \right\rangle$$-$we are just changing the basis. In the other case$-$where we act on the old state with the operator $\left(\sqrt{g'(\hat{x})}\right)^{-1}$$-$we are creating a new state $\left\lvert \phi \right\rangle$. Therefore, as a mathematical function, $$ \frac{1}{\sqrt{g'({x})}}\psi_{\hat{x}}(x) = \frac{1}{\sqrt{g'(g^{-1}(u))}}\psi_{\hat{x}}(g^{-1}(u)) $$ can represent either a new state or the representation of the old state in the new $u$-basis, and it all depends on what you are doing.


Thinking about the normalization is something of a red herring, because having a non-normalized state is not the same as changing the state: all multiples of a particular vector represent the same state as that vector (in the abstract Dirac state space, as in any vector space). In order to solve the ambiguity, you have to bring in some of the interpretational machinery of quantum mechanics, via either operators or probabilities or both.

  • $\begingroup$ Thank you so much. However, I had another confusion as I stated, that yeah I can absorb the measure into the ket, to interpret this as a new state. But I was talking about another thing. Can I absorb $g^{-1}$ function in $\langle g^{-1}(u)|\psi\rangle$ to get some other $\langle u|\phi\rangle$ ? $\endgroup$ Commented Nov 30, 2021 at 19:17
  • $\begingroup$ I'm not talking about the measure here. I can choose to absorb it into the ket, or to keep it separate. I'm talking about the fact that $\langle x|\psi\rangle = \langle g^{-1}(u)|\psi\rangle$. The latter one is some function of $u$. So technically we can think of it as the $u$ representation of some arbitrary ket. $\endgroup$ Commented Nov 30, 2021 at 19:20
  • $\begingroup$ Also, this going from $x$ to $u$, is not exactly a change of basis, in the standard sense, is it ? Normally I'd think of integral transform when I want to change basis, like from position to momentum. $\endgroup$ Commented Nov 30, 2021 at 19:21
  • $\begingroup$ Hello again! I see that I slightly misunderstood the question! I think that the answer to your question is basically in the post-script. It doesn't make sense to absorb the factor of $\sqrt{h(u)}$ into the abstract vector $|\psi\rangle$ and thereby call it a new state exactly because $\sqrt{h(u)}$ is a number, and multiplying a vector by a number yields the same state, although an unnormalized version. It think the rest of the post is important for clarification for how we talk about these things, how to notate them, but I'll rearrange things to put that point front and center. $\endgroup$
    – march
    Commented Nov 30, 2021 at 20:21
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    $\begingroup$ See the updated post! I added a "Part 2" to address the last question. Maybe that does it? $\endgroup$
    – march
    Commented Nov 30, 2021 at 21:21

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