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In quantum mechanics, we very often deal with ket vectors. Usually if two vectors belong to the same vector space, we can add them component wise. But my question is that is it permissible for ket vector? Like can I add $|x_1\rangle + |x_2\rangle$ to get $|x_1+x_2\rangle$? Cause I don't think both correspond to the same physical thing. Please enlighten me.

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    $\begingroup$ Related. $\endgroup$ Commented Sep 18, 2022 at 14:47
  • $\begingroup$ Is there any reason you did not accept an answer yet? $\endgroup$ Commented Mar 5, 2023 at 12:46

4 Answers 4

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You can certainly add any two kets together. However, assuming that you're referring to (generalized) position eigenvectors, the result is not $|x_1+x_2\rangle$.


Imagine that the Hilbert space is given by $\mathbb C^2$, the ordered pairs of complex numbers. Given the operator $$A := \pmatrix{1&0\\0&2}$$ I can define the states $$|1\rangle \equiv \pmatrix{1\\0} \qquad |2\rangle \equiv \pmatrix{0\\1}$$ where we have that $A|1\rangle = 1 |1\rangle$ and $A|2\rangle = 2|2\rangle$. I can certainly add $|1\rangle+|2\rangle$ if I wish; the resulting element of $\mathbb C^2$ would be $\pmatrix{1\\1}$. However, I can't say $|1\rangle+|2\rangle = |3\rangle$ (not least because the right-hand side has not even been defined).

In terms of (generalized) position eigenvectors with the Hilbert space $L^2(\mathbb R)$, $|x\rangle$ and $|y\rangle$ are eigenvectors of the position operator with eigenvalues $x$ and $y$, respectively. Crucially, note that $x$ and $y$ are labels for our vectors. While there is a state whose label is $x+y$, it is not given by $|x\rangle + |y\rangle$ for the same reason that $|1\rangle+|2\rangle \neq |3\rangle$ above.

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No, those are different vectors. The Hilbert space used here is a free vector space (See here) equipped with a canonical scalar product. (In general, there is a construction called free object (See here) in category theory, which is basically a mathematical structure like a group generated by a set.)

In the case of a free vector space, you take a field $\mathbb{K}$ and a set $X$ and then make the latter the basis of your vector space. That means the elements of this vector space are formal sums of the form: $$\sum_{x\in X}\lambda_xx$$ with $\forall x\in X\colon\lambda_x\in\mathbb{K}$. The free vector space itself is also often written $\mathbb{K}\otimes X$ for this reason. Of course, its dimension (the cardinality of a basis) is the cardinality of the set.

In quantum mechanics, you consider the free vector space generated by the underlying set of the vector space $\mathbb{R}^n$ (Note, that the new free vector space is not $\mathbb{R}^n$ again for it has infinite dimension while $\mathbb{R}^n$ has finite dimension) and consider all the basis vectors generating the free vector space to be orthogonal: $$\langle x|y\rangle=\delta(x-y).$$ $|x_1\rangle$, $|x_2\rangle$ and $|x_1+x_2\rangle$ are therefore three different and distinct basis vectors, that are pairwise orthogonal to each other and $|x_1\rangle+|x_2\rangle$ is a formal sum like in the equation above. Basically, since infinite dimensional spaces are hard to imagine, take $\mathbb{R}^3$ with the three basis vectors $e_1$, $e_2$ and $e_3=e_{1+2}$. I used $1+2=3$ here to highlight, it's really just about the name of the vector. We have: $$e_1+e_2 =\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \neq\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} =e_3=e_{1+2}.$$ Now fill this example up with infinitly many zeros and you have the same in a free vector space.

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To add to the other answer, some books define $|aA+bB\rangle$ as equal to $a|A\rangle+b|B\rangle$.The former is just a shorthand notation for the latter. In that notation, the $aA+bB$ written inside the ket is only a name for the ket.

But this notation can get confusing with position kets, like the other answer showed. $|x_1+x_2\rangle$ can either mean $|x_1\rangle+|x_2\rangle$ (if we're using the notation), or it can mean a position ket with eigenvalue $x_1+x_2$. I think the latter interpretation would be standard.

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  • $\begingroup$ Can you include examples of books using this horrible inconsistent notation? $\endgroup$
    – fqq
    Commented Sep 18, 2022 at 16:53
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I actually don't understand why do you think that $|x\rangle+|y\rangle=|x+y\rangle$. Yes, we can sum vectors in a vector space, in particular we can sum $|x\rangle+|y\rangle$ but there is absolutely nothing suggesting it should be $|x+y\rangle$. In fact, maybe one example clears this. Let me define a function depending on a parameter $x\in \mathbb{R}$: $$f_x(x')=\sin(x-x').$$

Now, we may obviously sum functions, so we may consider $f_x+f_y$. Is it the case that just because we can sum functions we have $f_{x+y}=f_{x}+f_y$? In this case this is obviously not true, as is the case for the position eigenstates.

On the other hand, perhaps the way to convince you that $|x+y\rangle$ and $|x\rangle+|y\rangle$ are different is to understand how they are used and to give an explicit counterexample. The position basis $|x\rangle$ is used to project states $|\psi\rangle$ to get the associated wavefunctions $\psi(x)=\langle x|\psi\rangle$ so that we may expand the state $$|\psi\rangle=\int dx \psi(x) |x\rangle\tag{1}.$$

In particular, the possible $\psi(x)$ in (1) are elements of $L^2(\mathbb{R})$, the space of square integrable functions on the real line.

Now, let $|\psi\rangle$ be an arbitrary state. Equality between $|x+y\rangle$ and $|x\rangle+|y\rangle$ would mean $$\psi(x+y)=\langle x+y|\psi\rangle=\langle x|\psi\rangle+\langle y|\psi\rangle=\psi(x)+\psi(y)\tag{2}.$$

In other words, a generic $\psi(x)$ would have to be linear and this is certainly not true. In fact, we may consider gaussian states by picking $\psi(x) = e^{-\alpha x^2}$. This is an element of $L^2(\mathbb{R})$, we have the associated $|\psi\rangle$ as given in (1) and (2) is simply not true for it.

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