Let {$|a_n\rangle$} and {$|b_n\rangle$} be two basis related by: $|b_n\rangle = \hat{U}|a_n\rangle \forall n$.

From my understanding then the unitary operator $\hat{U}$ only transforms the basis {$|a_n\rangle$} into {$|b_n\rangle$} (just like in 2D geometry having a rotation operator which changes the basis $\hat{x},\hat{y}$ to $\hat{r},\hat{\theta}$).

If there is an operator $\hat{\Omega}$, then its representation in basis {$|b_n\rangle$}: $$ \langle b_n|\Omega|b_m\rangle = \langle a_n| \hat{U}^\dagger\Omega\hat{U}|a_m\rangle $$ $$\Omega \to \hat{U}^\dagger\Omega\hat{U}$$

On the other hand, consider the following unitary transformation: $$|\psi\rangle = \Omega|\phi\rangle$$ $$\hat{U}|\psi\rangle = \hat{U}\Omega\hat{U}^\dagger\hat{U}|\phi\rangle$$ $$\Omega \to \hat{U}\Omega\hat{U}^\dagger$$

1)I am getting very confused by the difference between these, shouldn't the operator $\Omega$ transform in the same way?What is the difference between the two things I am doing?

  • $\begingroup$ See Hanting's comment. This is a directional confusion: $U|\psi\rangle$ would be going from $b$ basis to $a$ basis. For consistency with your first part, you would want to substitute $|\psi\rangle \rightarrow U^{\dagger}|\psi\rangle, |\phi\rangle \rightarrow U^{\dagger}|\phi\rangle$, and $\Omega \rightarrow U^{\dagger}\Omega U$ into your $|\psi\rangle = \Omega|\phi\rangle$, and then you have consitency. $\endgroup$
    – dsm
    Commented Mar 16, 2019 at 19:23
  • $\begingroup$ I don't think I am going from $b$ basis to $a$ basis; on contrary I believe that transformation corresponds to $a\to b$. Imagine I have an operator $\Omega$ with a known matrix representation in basis $a$. What I am doing is finding the corresponding representation in basis $b$ i.e. $a \to b$; which transforms as $\Omega \to U^\dagger \Omega U$. Isn't that correct? $\endgroup$ Commented Mar 16, 2019 at 20:52
  • $\begingroup$ In the first part yes, but lets look at the components of $\psi$ in the $a$ basis: $\langle a|\psi\rangle = \sum_{b}\langle a|b\rangle \langle b|\psi\rangle$, which if we use subscripts to note the basis is saying that $\psi_a = U\psi_b$. What you are wanting to look at in that second to last equation is $U^{\dagger}|\psi\rangle$. And note that this is all because in the first part you originally established your $U$ as going from $a$ to $b$. $\endgroup$
    – dsm
    Commented Mar 16, 2019 at 21:15

3 Answers 3


I suppose I'll formally write this up since there seems to still be some confusion. Let's firmly establish that our $U$ is a transformation from $a$ to $b$, that has it's representation in the $a$ basis as

$$\langle a_i |U|a_j\rangle = \langle a_i|b_j\rangle$$

Let's look at how the representation of $|\psi\rangle$ in the $a$ basis transforms when we go to the $b$ basis:

$$|\psi\rangle = \sum_{j}|b_j\rangle\langle b_j|\psi\rangle=\sum_{j}\sum_{i}|b_j\rangle\langle b_j|a_i\rangle\langle a_i|\psi\rangle$$

Now pick out the $k$'th component of $b$

$$\langle b_k|\psi\rangle = \sum_{j}\sum_{i}\delta_{kj}\langle b_j|a_i\rangle\langle a_i|\psi\rangle = \sum_{i}\langle b_k|a_i\rangle\langle a_i|\psi\rangle \\ = \sum_{i}(\langle a_i|b_k\rangle)^{\dagger}\langle a_i|\psi\rangle = \sum_{i}(\langle a_i|U|a_k\rangle)^{\dagger}\langle a_i|\psi\rangle.$$

Letting subscripts denote the basis (i.e. $|\psi\rangle_a \equiv \langle\vec{a}|\psi\rangle$, and likewise for $b$), we see that this is telling us $|\psi\rangle_b = U^{\dagger}|\psi\rangle_a$. Now, from $U|a_i\rangle=|b_i\rangle$ we know that $\Omega_b = U^{\dagger}\Omega_a U$, so lets check that everything is consistent with your $|\psi\rangle = \Omega|\phi\rangle$ when we hit it with $U^{\dagger}$. Keeping subscripts to denote the basis for absolute clarity:

$$|\psi\rangle_a = \Omega_a|\phi\rangle_a \to U^{\dagger}|\psi\rangle_a = U^{\dagger}\Omega_a|\phi\rangle_a$$

Looking at each side individually, we have

$$U^{\dagger}|\psi\rangle_a = |\psi\rangle_b \\ U^{\dagger}\Omega_a|\phi\rangle_a =U^{\dagger}\Omega_a U U^{\dagger}|\phi\rangle_a = \Omega_b |\phi\rangle_b,$$

which shows us that everything is nice and consistent:

$$ |\psi\rangle_a = \Omega_a|\phi\rangle_a \xrightarrow{U^{\dagger}} |\psi\rangle_b = \Omega_b|\phi\rangle_b $$

  • $\begingroup$ I just want to add something in case someone has the same confusion I had. The operator $\hat{U}$ transforms the basis $|a\rangle$ into the basis $|b\rangle$; a helpfull analogy would be a rotation in 2D; imagine that our new basis is a rotation by an angle $\theta$. If we are interested on how a 2D vector looks like from the perspective of the rotated basis, all we have to do is rotate the vector in the opposite direction that the basis rotated. That is why $|\psi\rangle_b$ is obtained by multiplying $|\psi\rangle_a$ with $\hat{U}^\dagger$. Please correct me if I'm wrong. $\endgroup$ Commented Mar 17, 2019 at 19:02
  • $\begingroup$ That is correct. Not to add any confusion, but I will add that although $U$ has been constructed to take $a$ to $b$, it likewise can take $b$ to $a$: $|\psi\rangle_a = U|\psi\rangle_b$ and $\Omega_a = U\Omega_b U^{\dagger}$. This is just from hitting our $|\psi\rangle_b = U^{\dagger}|\psi\rangle_a$ with a $U$ and the left, and sandwiching $\Omega_b = U^{\dagger}\Omega_a U$ between a $U$ on the left and $U^{\dagger}$ on the right. You can then carry over your rotation analogy to see why these also make sense. $\endgroup$
    – dsm
    Commented Mar 17, 2019 at 19:31

In the first example, you're taking $|b_i\rangle$ into $|a_i\rangle$, while in the second, you're doing the opposite. Explicitly, if $$|\psi\rangle = \sum_n b_n|b_n\rangle$$ then $$\hat U|\psi\rangle = \sum_n b_n\hat U|b_n\rangle.$$

But we don't know anything about $U|b_n\rangle$! To go from $|b_i\rangle$ into $|a_i\rangle$, we what we need is $\hat U^\dagger$. Then the calculation will come out as $\Omega \mapsto \hat U^\dagger \Omega \hat U$.

  • $\begingroup$ Could you clarify why am I going from $|b\rangle$ into $|a\rangle$ in my first example? The way I see it is: I know the matrix representation of $\Omega$ in basis $|a\rangle$, then I find the matrix representation in $|b\rangle$ which is $U^\dagger \Omega U$ i.e. I went from $|a\rangle$ to $|b\rangle$. Isn't that correct? $\endgroup$ Commented Mar 16, 2019 at 20:55
  • $\begingroup$ Well, you can see from the matrix elements that you've gone from $\langle b_n|\Omega|b_m\rangle$ to $\langle a_n| \hat{U}^\dagger\Omega\hat{U}|a_m\rangle$. The left side has $|b_n\rangle$, the right side has $|a_n\rangle$ $\endgroup$
    – cxx
    Commented Mar 17, 2019 at 0:01
  • $\begingroup$ Yes, in the left hand side I have $|b\rangle$ which means that I have obtained it from the rhs $|a\rangle$ i.e. $|a\rangle \to |b\rangle$. Sorry if it is too obvious, but I just can't see it. In the comments on the question someone agrees that I went from $|a\rangle \to |b\rangle$. $\endgroup$ Commented Mar 17, 2019 at 0:11

$\let\Om=\Omega \let\dag=\dagger \def\ket#1{|#1\rangle} \def\bra#1{\langle#1|} \def\braket#1#2{\langle#1|#2\rangle} \def\mxelm#1#2#3{\bra#1\,#2\,\ket#3}$

As you'll see, I'll make some change of notations I can't explain for brevity. In your first definition you're asking for an operator $\Om'$ which in $a$-rep has the same matrix elements $\Om$ has in $b$-rep: $$\mxelm{{a,m}}{\Om'}{{a,n}} = \mxelm{{b,m}}\Om{{b,n}}$$ $$\Om' = U^\dag\,\Om\,U.$$

Afterwards you assume some operator $\Om$ sends ket $\xi$ into $\eta$: $$\ket\eta = \Om\,\ket\xi.$$ Then you define $\ket{\xi'}$, $\ket{\eta'}$ such that they have in $b$-rep the same components $\ket\xi$, $\ket\eta$ have in $a$-rep: $$\braket{b,n}{\xi'} = \braket{a,n}\xi$$ $$\mxelm{{a,n}}{U^\dag}{{\xi'}} = \braket{a,n}\xi$$ $$\ket{\xi'} = U\,\ket\xi \qquad \ket{\eta'} = U\,\ket\eta.$$ Lastly you look for $\Om'$ sending $\ket{\xi'}$ into $\ket{\eta'}$: $$\Om'\,U\,\ket\xi = U\,\ket\eta = U\,\Om\,\ket\xi.$$ If this is to hold for all $\ket\xi$ then $$\Om' = U\,\Om\,U^\dag.$$

It should be clear that in your former part you transformed an operator, whereas in the latter you transformed kets under the opposite requirement.


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