My 1D harmonic oscillator of mass $m$ is in a potential $V(x) = \frac{m\omega^2x^2}{2}$. Here's its state: $\left|\psi\right> = Ne^{\lambda a^\dagger}\left|0\right>$, where $\lambda$ is given, $\lambda \in \mathcal{C}$, and $a^\dagger$ is a ladder operator, $a^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} - \frac{i}{m\omega}\hat{p}\right)$.

I'm asked to normalize it (find $N$), but I don't understand what's the effect of having an $a^\dagger$ in the exponent, would that mean that my state decays in phase space? Or is just a mathematic exercise with no practical physical meaning?

UPD: I don't question the existence of such operator, or mathematical ways to compute $N$. This question is about physical picture of such operator.

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    $\begingroup$ I think you should use that $\exp(A) = \sum_n \frac{A^n}{n!}$ and that $a^\dagger |n\rangle = \sqrt{n+1} |n+1\rangle$. $\endgroup$
    – Jeroen
    Mar 6, 2017 at 11:52
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    $\begingroup$ The squeezed coherent states $e^{\lambda a^* -\bar{\lambda} a}\lvert 0\rangle $ are very important in quantum physics. For example, they are the vectors with minimal uncertainty and they are often used to model lasing in quantum optics. The corresponding exponential operator $e^{\lambda a^* -\bar{\lambda} a}$ is sometimes called the Weyl operator, and it is related to the canonical commutation relations in exponentiated form (and appears in many applications, both physical and mathematical). $\endgroup$
    – yuggib
    Mar 6, 2017 at 12:36
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    $\begingroup$ The physical picture is pretty simple: this is a so-called coherent state, or a Gaussian wave-packet peaked at some point $(x, p)$ of the phase space. Coherent states are as close to the classical description as you can get in the quantum theory :) $\endgroup$ Mar 6, 2017 at 14:03
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    $\begingroup$ @OscarMozhajskii Just see en.wikipedia.org/wiki/Coherent_states, especially Secs.2 and 3. It's a coherent state, but not a squeezed coherent state. $\endgroup$
    – udrv
    Mar 6, 2017 at 20:00

2 Answers 2


The exponential of the creation operator has a lot of physical meaning, the state is very important.

There are several classes of wisdom about it, including

  • how it's calculated
  • what algebraic properties the exponential or its action on a state obeys
  • what is the physical importance or interpretation of these things

The exponential of an operator may be calculated e.g. using Taylor series $$ \exp(X) = \sum_{n=0}^\infty \frac{X^n}{n!}$$ This holds even if $X$ is an operator, e.g. a creation operator. The power $X^n$ is calculated from the usual matrix multiplication (with itself). An alternative calculation is $$ \exp(X) = \lim_{M\to\infty} (1+X/M)^M $$ which resembles the compound interest in banking.

More general functions of an operator may be found by diagonalizing the operator as $$ X = U D U^{-1}$$ and then $$ f(X) = U f(D) U^{-1}$$ which holds because $U^{-1}$ and $U$ cancel in the power $X^N$, and $f(D)$, a function of a diagonal matrix, is calculated by applying the function $f$ on each diagonal matrix entry.

At any rate, the operator $\exp(\lambda a^\dagger)$ perfectly exists, is calculable, and is a well-defined linear operator on the same Hilbert space.

The normalization $N$ may be calculated if you evaluate the squared norm Insert e.g. the "limit" form of the exponential for the exponential above. You will have $$\langle 0 | (1+\lambda a/N)^N (1+\lambda a^\dagger)^N |0 \rangle$$ in the limit $N\to\infty$. How much is the matrix element above? If you move all the factors $(1+\lambda a/N)^N$ to the right, next to the ket $|0\rangle$, their action on the ket will be simply $1$ and you may forget about those. Similarly with the action of the Hermitian conjugate factors on the bra-vector.

To permute them, you need to exchange the order of the $(1+\lambda a/N)$ and $(1+\lambda a^\dagger/N)$ factors. There are $N^2$ pairs of such factors that you need to exchange and you use $$ (1+\lambda a/N) (1+\lambda a^\dagger/N) = (1+\lambda a^\dagger/N) (1+\lambda a/N) (1+\lambda^2 [a,a^\dagger]/N^2) $$ where the last multiplicative correction just adjusts the terms of order $1/N^2$, in a $1/N$ expansion. Also, $[a,a^\dagger]=1$. When $N^2$ such permutations are made, the $(1+\lambda^2 [a,a^\dagger]/N^2)$ factors multiplicatively combine to $$\exp(\lambda^2) $$ by the same limit-based definition of the exponential – the limit $N^2\to \infty$ is the same as $N\to\infty$ and the $1/N^2$ factor cancels.

So the norm is $\exp(\lambda^2)$ and to guarantee the normalization to one, as you need it, you need $$ N_{\rm yours} = \exp(-|\lambda|^2/2)$$ I added the absolute value because $\lambda$ may be complex. In that case, one-half of the $\lambda$'s above, all the bra-style ones, need to be complex conjugated and we end up with the absolute value.

The exponential of $\lambda a^\dagger$ acting on the vacuum has the effect of moving the Gaussian wave function in the $x$ and $p$ direction of the phase space – according to the real and imaginary part of $\lambda$. The state is known as a coherent state and you may prove that it's an eigenstate of $a$. Just like $a$ itself annihilates $|0\rangle$, its action on the coherent state is still simple, the latter is an eigenstate, and the eigenvalue is $\lambda$ or so (or perhaps its complex conjugate, or it over two etc., find it yourself).

There are lots of calculations to be made here and they're important and may be done in various ways. But I want to stress the more general point. The exponentials of all similar operators are meaningful and very important in quantum mechanics and lots of important laws and states and objects are indeed written in this way. It is not just some silly mathematics where objects are unnaturally substituted where they don't belong. They certainly do belong there and many important if not elementary objects have a similar mathematical form.

  • $\begingroup$ Thanks for your response. This question is not about exponential operators. I like the one-before-the-last paragraph of your answer. Would you say it's similar to the evolution operator, but operates on the phase space with $\lambda$ being direction parameter? $\endgroup$ Mar 6, 2017 at 12:19
  • $\begingroup$ It is a generalization of an evolution operator - in both directions of the phase space - but it is not unitary. It is not unitary even after you multiply it by the normalization factor. The problem is that it moves the Gaussian ground state in the x,p directions and the normalization factor you calculate is enough for that. However, when you're moving excited states, you need a different normalization constant. So when acting on "pieces", it's just like an evolution operator, but as a whole, it's rather far from a unitary operator. $\endgroup$ Mar 8, 2017 at 5:03

I did a quick calculation and (using the earlier mentioned relations) I found the following (beware, I may have been sloppy)

$|\psi\rangle=N\sum_n\frac{\sqrt{(n+1)}\lambda^n}{\sqrt{n!}}|n+1\rangle$. Inserting the eigenfunctions of the Hamiltonian $$ \psi_n(x) = N_n \cdot e^{ - \frac{m\omega x^2}{2 \hbar}} \cdot H_n\left(\sqrt{\frac{m\omega}{\hbar}} x \right) $$ where the normalization constants are $$ \frac{1}{\sqrt{2^n\,n!}} \cdot \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} $$ Inserting this in the above sum we find $$ \begin{align} |\psi\rangle=&N\sum_n\frac{\lambda^n}{2^{n/2} n!} \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} e^{ - \frac{m\omega x^2}{2 \hbar}} \cdot H_{n+1}\left(\sqrt{\frac{m\omega}{\hbar}} x \right) \\ =& N e^{\lambda/\sqrt{2}} \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} e^{ - \frac{m\omega x^2}{2 \hbar}} \cdot H_{n+1}\left(\sqrt{\frac{m\omega}{\hbar}} x \right) \end{align} $$ So your normalization constant seems to be is $$ N = e^{-\lambda/\sqrt{2}} \left(\frac{m\omega}{\pi \hbar}\right)^{-1/4} $$


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